Differential Equations 9788131770375, 9789332511644, 8131770370

Table of contents :
Cover......Page 1
Contents......Page 4
Preface......Page 12
About the Author......Page 13
1.1.1 Differential Equation......Page 14
1.2.1 Formation of a Differential Equation......Page 17
1.2.2 Solution of a Differential Equation......Page 25
Exercise 1.1......Page 26
2.1 First order and first degree differential equations......Page 30
2.1.1 Variable Separable Equation......Page 31
Exercise 2.1......Page 36
2.1.2 Homogeneous Equations......Page 38
Exercise 2.2......Page 43
2.1.3 Non-homogeneous Equations......Page 45
Exercise 2.3......Page 59
2.1.4 Exact Equations......Page 61
Exercise 2.4......Page 68
2.1.5 Inexact Equation—Reducible to Exact Equation by Integrating Factors......Page 69
Exercise 2.5......Page 73
Exercise 2.6......Page 83
2.1.6 Linear Equations......Page 85
Exercise 2.7......Page 95
2.1.7 Bernoulli’s Equation......Page 97
Exercise 2.8......Page 101
2.2 Applications of ordinary differential equations......Page 103
Exercise 2.9......Page 107
2.2.1 Geometrical Applications......Page 108
Exercise 2.10......Page 115
3.1 Introduction......Page 116
3.1.1 Linear Differential Equations of the Second Order......Page 117
3.1.2 Homogeneous Equations—Superposition or Linearity Principle......Page 119
3.1.3 Fundamental Theorem for the Homogeneous Equation......Page 120
3.1.4 Initial Value Problem (IVP)......Page 121
3.1.5 Linear Dependence and Linear Independence of Solutions......Page 123
3.1.7 Second Order Linear Homogeneous Equations with Constant Coefficients......Page 124
Exercise 3.1......Page 128
3.1.8 Higher Order Linear Equations......Page 129
3.1.9 Linearly Independent (L.I.) Solutions......Page 130
3.1.10 Exponential Shift......Page 132
Exercise 3.2......Page 137
3.1.11 Inverse operator D−1 or 1D......Page 139
3.1.12 General Method for Finding the P. I.......Page 140
Exercise 3.3......Page 142
3.2 General solution of linear equation f (D) y = Q(x) 3-29......Page 144
Exercise 2.4......Page 150
3.2.1 Short Methods for Finding the Particular Integrals in Special Cases......Page 151
Exercise 3.5......Page 157
Exercise 3.6......Page 165
Exercise 3.7......Page 168
Exercise 3.8......Page 171
Exercise 3.9......Page 176
3.2.2 Linear Equations with Variable Coefficients— Euler–Cauchy Equations (Equidimensional Equations)......Page 178
Exercise 3.10......Page 184
3.2.3 Legendre’s Linear Equation......Page 186
Exercise 3.11......Page 188
3.2.4 Method of Variation of Parameters......Page 189
Exercise 3.12......Page 196
3.2.5 Systems of Simultaneous Linear Differential Equations with Constant Coefficients......Page 197
Exercise 3.13......Page 201
4.1 Equations solvable for p......Page 204
Exercise 4.1......Page 213
4.2 Equations solvable for y......Page 214
Exercise 4.2......Page 222
4.3 Equations solvable for x......Page 223
Exercise 4.3......Page 229
Chapter 5: Linear Equation of the Second Order with Variable Coefficients......Page 230
5.1 To find the integral in C.F. by inspection, i.e. to find a solution of......Page 232
Exercise 5.1......Page 245
5.2 General solution of R by changing the dependent variable and removing the first derivative (Reduction to normal form)......Page 247
Exercise 5.2......Page 252
5.3 General solution of by changing......Page 254
Exercise 5.3......Page 261
6.1.2 Power Series Method of Solution of Linear Differential Equations......Page 264
6.1.3 Existence of Series Solutions: Method of Frobenius......Page 265
6.1.4 Legendre Functions......Page 267
6.1.5 Legendre Polynomials Pa(x)......Page 269
6.1.6 Generating Function for Legendre Polynomials Pn(x)......Page 275
6.1.7 Recurrence Relations of Legendre Functions......Page 276
6.1.9 Orthogonality of Legendre Polynomials Pn(x)......Page 280
6.1.11 Christoffel’s Expansion......Page 283
6.1.12 Christoffel’s Summation Formula......Page 284
6.1.13 Laplace’s First Integral for (x)......Page 285
6.1.14 Laplace’s Second Integral for (x)......Page 286
6.1.15 Expansion of f(x) in a Series of Legendre Polynomials......Page 287
Exercise 6.1......Page 291
6.2.2 Bessel Functions......Page 292
6.2.3 Bessel Functions of Non-integral Order p: Jp(x) and J-p(x)......Page 294
6.2.4 Bessel Functions of Order Zero and One: J0(x), J1(x)......Page 295
6.2.5 Bessel Function of Second Kind of Order Zero Y0(x)......Page 296
6.2.6 Bessel Functions of Integral Order: Linear Dependence of Jn(x) and J-n(x)......Page 297
6.2.8 Generating Functions for Bessel Functions......Page 298
6.2.9 Recurrence Relations of Bessel Functions......Page 300
6.2.10 Bessel’s Functions of Half-integral Order......Page 302
6.2.11 Differential Equation Reducible to Bessel’s Equation......Page 304
6.2.12 Orthogonality......Page 305
6.2.13 Integrals of Bessel Functions......Page 308
6.2.14 Expansion of Sine and Cosine in Terms of Bessel Functions......Page 309
Exercise 6.2......Page 315
6.3 Chebyshev polynomials......Page 316
Exercise 6.3......Page 329
7.2.1 Laplace Transform......Page 330
7.3 Fourier integral theorem......Page 331
7.3.1 Fourier Sine and Cosine Integrals (FSI’s and FCI’s)......Page 332
7.4.1 Fourier Integral Representation of a Function......Page 333
7.5 Fourier transform of f (x)......Page 334
7.5.1 Fourier Sine Transform (FST) and Fourier Cosine Transform (FCT)......Page 335
7.6 Finite Fourier sine transform and finite Fourier cosine transform (FFCT)......Page 336
7.6.1 FT, FST and FCT Alternative definitions......Page 337
7.7.2 Convolution Theorem......Page 338
7.8.1 Linearity Property......Page 339
7.8.3. Shifting Property......Page 340
7.8.4 Modulation Theorem......Page 341
Exercise 7.1......Page 358
7.9 Parseval’s identity for Fourier transforms......Page 360
7.10 Parseval’s identities for Fourier sine and cosine transforms......Page 361
Exercise 7.2......Page 364
8.2.1 Order......Page 366
8.2.3 Homogeneity......Page 367
8.3 Origin of partial differential equation......Page 368
8.4 Formation of partial differential equation by elimination of two arbitrary constants......Page 369
Exercise 8.1......Page 373
8.5 Formation of partial differential equations by elimination of arbitrary functions......Page 374
Exercise 8.2......Page 379
8.6.1 Linear Equation......Page 380
8.7 Classification of solutions of first-order partial differential equation......Page 381
8.7.2 General Integral......Page 382
8.8 Equations solvable by direct integration......Page 383
Exercise 8.3......Page 386
8.9 Quasi-linear equations of first order......Page 387
8.10.2 Two Variables are Separable......Page 389
8.10.3 Method of Multipliers......Page 391
Exercise 8.4......Page 401
8.11 Non-linear equations of first order......Page 403
Exercise 8.5......Page 412
8.12 Euler’s method of separation of variables......Page 413
Exercise 8.6......Page 418
8.13.1 Introduction......Page 419
8.13.2 Classification of Equations......Page 420
8.13.4 Solution of One-dimensional Heat Equation (or diffusion equation)......Page 421
Exercise 8.7......Page 435
8. 13.6 Vibrating String with Zero Initial Velocity......Page 437
8.13.7 Vibrating String with Given Initial Velocity and Zero Initial Displacement......Page 443
8.13.8 Vibrating String with Initial Displacement and Initial Velocity......Page 450
Exercise 8.8......Page 452
8.13.9 Laplace’s equation or potential equation or two-dimensional steady-state heat flow equation equation......Page 453
Exercise 8.9......Page 461
Exercise 8.10......Page 469
Index......Page 470

Citation preview

Differential Equations

E. Rukmangadachari Professor of Mathematics Department of Humanities and Sciences Malla Reddy Engineering College Secunderabad

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Copyright © 2012 Dorling Kindersley (India) Pvt. Ltd. Licensees of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material present in this eBook at any time. ISBN 9788131770375 eISBN 9789332511644 Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India Registered Office: 11 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India

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Contents

Preface About the Author

1. Formation of a Differential Equation 1.1 1.2

xi xii

1-1

Introduction 1-1 1.1.1 Differential equation 1-1 Differential equations 1-4 1.2.1 Formation of a differential equation 1-4 1.2.2 Solution of a differential equation 1-12 Exercise 1.1 1-13

2. Differential Equations of First Order and First Degree 2-1 2.1

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First order and first degree differential equations 2-1 2.1.1 Variable separable equation 2-2 Exercise 2.1 2-7 2.1.2 Homogeneous equations 2-9 Exercise 2.2 2-14 2.1.3 Non-homogeneous equations 2-16 Exercise 2.3 2-30 2.1.4 Exact equations 2-32 Exercise 2.4 2-39 2.1.5 Inexact equation—Reducible to exact equation by integrating factors 2-40 Exercise 2.5 2-44 Exercise 2.6 2-54 2.1.6 Linear equations 2-56 Exercise 2.7 2-66 2.1.7 Bernoulli’s equation 2-68 Exercise 2.8 2-72

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iv

Contents

2.2

Applications of ordinary differential equations Exercise 2.9 2-78 2.2.1 Geometrical applications 2-79 Exercise 2.10 2-86

2-74

3. Linear Differential Equations with Constant Coefficients

3-1

3.1

Introduction 3-1 3.1.1 Linear differential equations of the second order 3-2 3.1.2 Homogeneous equations—superposition or linearity principle 3-4 3.1.3 Fundamental theorem for the homogeneous equation 3-5 3.1.4 Initial value problem (IVP) 3-6 3.1.5 Linear dependence and linear independence of solutions 3-8 3.1.6 General solution, basis and particular solution 3-9 3.1.7 Second order linear homogeneous equations with constant coefficients 3-9 Exercise 3.1 3-13 3.1.8 Higher order linear equations 3-14 3.1.9 Linearly independent (L.I.) solutions 3-15 3.1.10 Exponential shift 3-17 Exercise 3.2 3-22 3.1.11 Inverse operator D−1 or D1 3-24 3.1.12 General method for finding the P. I. 3-25 Exercise 3.3 3-27 3.2 General solution of linear equation f ( D ) y = Q ( x ) 3-29 Exercise 2.4 3-35 3.2.1 Short methods for finding the particular integrals in special cases 3-36 Exercise 3.5 3-42

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Contents

3.2.2

3.2.3 3.2.4 3.2.5

Exercise 3.6 3-50 Exercise 3.7 3-53 Exercise 3.8 3-56 Exercise 3.9 3-61 Linear equations with variable coefficients— Euler–Cauchy equations (Equidimensional equations) 3-63 Exercise 3.10 3-69 Legendre’s linear equation 3-71 Exercise 3.11 3-73 Method of variation of parameters 3-74 Exercise 3.12 3-81 Systems of simultaneous linear differential equations with constant coefficients 3-82 Exercise 3.13 3-86

4. Differential Equations of the First Order but not of the First Degree 4.1 4.2 4.3

5.2

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4-1

Equations solvable for p 4-1 Exercise 4.1 4-10 Equations solvable for y 4-11 Exercise 4.2 4-19 Equations solvable for x 4-20 Exercise 4.3 4-26

5. Linear Equation of the Second Order with Variable Coefficients 5.1

v

5-1

To find the integral in C.F. by inspection, i.e. to find d2 y dy a solution of + P + Qy = 0 5-3 2 dx dx Exercise 5.1 5-16 d2 y dy General solution of + P + Qy = R by changing 2 dx dx the dependent variable and removing the first derivative (Reduction to normal form) 5-18 Exercise 5.2 5-23

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vi

Contents

5.3

d 2 y dy P + Qy = R by changing dx 2 dx the independent variable 5-25 Exercise 5.3 5-32 General solution of

6. Integration in Series: Legendre, Bessel and Chebyshev Functions

6-1

6.1

Legendre functions 6-1 6.1.1 Introduction 6-1 6.1.2 Power series method of solution of linear differential equations 6-1 6.1.3 Existence of series solutions: Method of Frobenius 6-2 6.1.4 Legendre functions 6-4 6.1.5 Legendre polynomials Pn(x) 6-6 6.1.6 Generating function for Legendre polynomials Pn(x) 6-12 6.1.7 Recurrence relations of Legendre functions 6-13 6.1.8 Orthogonality of functions 6-17 6.1.9 Orthogonality of Legendre polynomials Pn(x) 6-17 6.1.10 Betrami’s result 6-20 6.1.11 Christoffel’s expansion 6-20 6.1.12 Christoffel’s summation formula 6-21 6.1.13 Laplace’s first integral for Pn(x) 6-22 6.1.14 Laplace’s second integral for Pn(x) 6-23 6.1.15 Expansion of f (x) in a series of Legendre polynomials 6-24 Exercise 6.1 6-28 6.2 Bessel functions 6-29 6.2.1 Introduction 6-29 6.2.2 Bessel functions 6-29

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Contents

vii

6.2.3 Bessel functions of non-integral order p: Jp(x) and J−p(x) 6-31 6.2.4 Bessel functions of order zero and one: J0(x), J1(x) 6-32 6.2.5 Bessel function of second kind of order zero Y0(x) 6-33 6.2.6 Bessel functions of integral order: Linear dependence of Jn(x) and J−n(x) 6-34 6.2.7 Bessel functions of the second kind of order n: Yn(x): Determination of second solution Yn(x) by the method of variation of parameters 6-35 6.2.8 Generating functions for Bessel functions 6-35 6.2.9 Recurrence relations of Bessel functions 6-37 6.2.10 Bessel’s functions of half-integral order 6-39 6.2.11 Differential equation reducible to Bessel’s equation 6-41 6.2.12 Orthogonality 6-42 6.2.13 Integrals of Bessel functions 6-45 6.2.14 Expansion of sine and cosine in terms of Bessel functions 6-46 Exercise 6.2 6-52 6.3 Chebyshev polynomials 6-53 Exercise 6.3 6-66

7. Fourier Integral Transforms 7.1 7.2

7.3

7.4

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7-1

Introduction 7-1 Integral transforms 7-1 7.2.1 Laplace transform 7-1 7.2.2 Fourier transform 7-2 Fourier integral theorem 7-2 7.3.1 Fourier sine and cosine integrals (FSI’s and FCI’s) 7-3 Fourier integral in complex form 7-4

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viii

7.5

7.6

7.7

7.8

7.9 7.10

Contents

7.4.1 Fourier integral representation of a function 7-4 Fourier transform of f (x) 7-5 7.5.1 Fourier sine transform (FST) and Fourier cosine transform (FCT) 7-6 Finite Fourier sine transform and finite Fourier cosine transform (FFCT) 7-7 7.6.1 FT, FST and FCT alternative definitions 7-8 Convolution theorem for Fourier transforms 7-9 7.7.1 Convolution 7-9 7.7.2 Convolution theorem 7-9 7.7.3 Relation between Laplace and Fourier transforms 7-10 Properties of Fourier transform 7-10 7.8.1 Linearity property 7-10 7.8.2 Change of scale property or damping rule 7-11 7.8.3 Shifting property 7-11 7.8.4 Modulation theorem 7-12 Exercise 7.1 7-29 Parseval’s identity for Fourier transforms 7-31 Parseval’s identities for Fourier sine and cosine transforms 7-32 Exercise 7.2 7-35

8. Partial Differential Equations 8.1 8.2

8.3 8.4

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8-1

Introduction 8-1 Order, linearity and homogeneity of a partial differential equation 8-1 8.2.1 Order 8-1 8.2.2 Linearity 8-2 8.2.3 Homogeneity 8-2 Origin of partial differential equation 8-3 Formation of partial differential equation by elimination of two arbitrary constants 8-4 Exercise 8.1 8-8

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Contents

8.5

8.6

8.7

8.8 8.9 8.10

8.11 8.12 8.13

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ix

Formation of partial differential equations by elimination of arbitrary functions 8-9 Exercise 8.2 8-14 Classification of first-order partial differential equations 8-15 8.6.1 Linear equation 8-15 8.6.2 Semi-linear equation 8-16 8.6.3 Quasi-linear equation 8-16 8.6.4 Non-linear equation 8-16 Classification of solutions of first-order partial differential equation 8-16 8.7.1 Complete integral 8-17 8.7.2 General integral 8-17 8.7.3 Particular integral 8-18 8.7.4 Singular integral 8-18 Equations solvable by direct integration 8-18 Exercise 8.3 8-21 Quasi-linear equations of first order 8-22 Solution of linear, semi-linear and quasi-linear equations 8-24 8.10.1 All the variables are separable 8-24 8.10.2 Two variables are separable 8-24 8.10.3 Method of multipliers 8-26 Exercise 8.4 8-36 Non-linear equations of first order 8-38 Exercise 8.5 8-47 Euler’s method of separation of variables 8-48 Exercise 8.6 8-53 Classification of second-order partial differential equations 8-54 8.13.1 Introduction 8-54 8.13.2 Classification of equations 8-55 8.13.3 Initial and boundary value problems and their solution 8-56

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Contents

8.13.4 Solution of one-dimensional heat equation (or diffusion equation) 8-56 Exercise 8.7 8-70 8.13.5 One-dimensional wave equation 8-72 8.13.6 Vibrating string with zero initial velocity 8-72 8.13.7 Vibrating string with given initial velocity and zero initial displacement 8-78 8.13.8 Vibrating string with initial displacement and initial velocity 8-85 Exercise 8.8 8-87 8.13.9 Laplace’s equation or potential equation or two-dimensional steady-state heat flow equation 8-88 Exercise 8.9 8-96 Exercise 8.10 8-104 Index

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I-1

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Preface

I am pleased to present this book on Differential Equations to the undergraduate students of all Indian universities. This book has been written in a simple, lucid and easy-to-understand manner. The concepts have been discussed with focus on clarity and coherence, supported by illustrations for better comprehension. Adequate examples have been worked out in the book to enable students understand the fundamentals and the principles governing each topic. The exercises given at the end of each chapter—with answers and hints wherever necessary—provide students an insight into the methods of solving the problems with ingenuity. Suggestions for the improvement of the book are welcome and will be gratefully acknowledged.

Acknowledgements I express my deep sense of gratitude to Sri Ch. Malla Reddy, Chairman, and Sri Ch. Mahender Reddy, Secretary, Malla Reddy Group of Institutions (MRGI), whose patronage has given me an opportunity to write this book. I am also thankful to Prof. R. Madan Mohan, Director (Academics); Col G. Ram Reddy, Director (Administration), MRGI; and Dr R. Somanatham, Principal, Malla Reddy Engineering College, Secunderabad, for their kindness, guidance, and encouragement. E. RUKMANGADACHARI

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About the Author

E. Rukmangadachari is former head of Computer Science and Engineering as well as Humanities and Sciences at Malla Reddy Engineering College, Secunderabad. Earlier, he was a reader in Mathematics (PG course) at Government College, Rajahmundry. He is an M.A. from Osmania University, Hyderabad, and an M.Phil. and Ph.D. degree holder from Sri Venkateswara University, Tirupathi. A recipient of the Andhra Pradesh State Meritorious Teachers’ Award in 1981, Professor Rukmangadachari has published over 40 research papers in national and international journals. With a rich repertoire of over 45 years’ experience in teaching mathematics to undergraduate, postgraduate and engineering students, he was the vice-president of the Andhra Pradesh Society for Mathematical Sciences. An ace planner with fine managerial skills, he was the organising secretary for the conduct of the 17th Congress of the Andhra Pradesh Society for Mathematical Sciences, Hyderabad.

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1 Formation of a Differential Equation 1.1

INTRODUCTION

The general laws of science, engineering, medicine, social sciences, population dynamics and the like where the rate of change of quantities is involved are usually modelled as differential equations. Formation of differential equations, solution and interpretation of the results are of practical interest, especially for engineers and physicists, as they deal with many engineering and physical problems. So, we study in this chapter, methods of solution of ordinary differential equations of the first order and first degree, their applications to Newton’s law of cooling the law of natural growth and decay, and orthogonal trajectories. Further, methods of solution of linear differential equations of second and higher order with constant coefficients are also considered in this chapter.

1.1.1

Differential Equation

An equation expressing a relation between functions, their derivatives and the variables is called a differential equation. Differential equations are classified into (1) ordinary and (2) partial differential equations.

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1-2

Differential Equations

Ordinary differential equation A differential equation containing derivatives of a function (or functions) of a single variable is called an ordinary differential equation. Example 1.1.1

(order, degree)

d2 y =0 dx 2 dy a 2. y = x + dx dy dx 2 d y + m2 y = 0 3. dx 2 1.

(2,1) (1,2)

(2,1)

2

d 2 y ⎛ dy ⎞ 2 4. y 2 − ⎜⎝ ⎟⎠ = y log y dx dx ⎡ ⎛ dy ⎞ 2 ⎤ 5. ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ dx ⎠ ⎥⎦ 6. y = x

32

=a

d2 y dx 2

dy ⎛ dy ⎞ + a 1+ ⎜ ⎟ ⎝ dx ⎠ dx

(2,2) 2

dy ⎞ ⎛ dx 7. ⎜ y + x ⎟ ( a − t ) + xy = 0 ⎝ dt dt ⎠

8.

(2,1)

d2x − k 2 y = a cos pt dt 2 d2 y + k 2 x = a sin pt dt 2

(1,2) (1,1)

(2,1)

Partial differential equation A differential equation containing partial derivatives of a function (or functions) of more than one variable, with each derivative referring to one of the variables is called a partial differential equation. Example 1.1.2 ∂ z ∂ z + =0 ∂x 2 ∂y 2 2

1.

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(order, degree)

2

(2,1)

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Formation of a Differential Equation

2.

∂2 z =0 ∂x ∂y

1-3

(2,1) 2

⎛ ∂z ⎞ ⎛ ∂z ⎞ 3. ⎜ ⎟ + ⎜ ⎟ = 0 ⎝ ∂x ⎠ ⎝ ∂y ⎠

(1,2)

∂z ∂z + xy = nxz ∂x ∂y

(1,1)

2

4. y 2 5. x

∂2u ∂2u + y = nu ∂x 2 ∂y 2

(2,1)

6. x

∂z ∂z +y =z ∂x ∂y

(1,1)

With each differential equation, we associate a pair of positive integers (n, m) called the order and degree, respectively, of the differential equation. Order of a differential equation The order of a differential equation is the order of the highest derivative appearing in it. The general form of an nth order ordinary differential equation in variables x and y is F ( x, y, y ′, y ′′, , y ( n ) = 0) where y ( n ) =

dn y . dx n

Degree of a differential equation The degree of a differential equation is the degree or power of the highest derivative, when the equation is freed from radicals and fractions in respect of the derivatives, i.e., when the expression F ( x, y, y ′, y ′′, , y ( n ) ) is written as a rational integral algebraic expression in y ( n ) . If F cannot be expressed in this manner then the degree of the differential equation is not defined. The order and degree of the differential equations given above are indicated against each of them.

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1-4

Differential Equations

We can also classify differential equations as linear and non-linear. A differential equation is called linear if the sum of the powers of the function and its derivatives in each term is at most equal to unity, and otherwise, non-linear. Example 1.1.3 1. The following differential equations are linear. (i) a0 ( x )

d2 y dy + a1 ( x ) + a2 ( x ) y = b( x ) 2 dx dx

d2 y + m2 y = 0 dx 2 2. The following differential equations are non-linear: (ii)

2

⎛ dy ⎞ (i) ⎜ ⎟ = x 2 y ⎝ dx ⎠

1.2

(ii)

dy = tan y dx

DIFFERENTIAL EQUATIONS

Differential equations arise in many engineering and physical problems. The approach of an engineer to the study of differential equations has to be practical, and so, it consists of (1) formation of a differential equation from the physical conditions, called modelling; (2) solution of a differential equation under the initial/boundary conditions; and (3) the physical interpretation of the results.

1.2.1

Formation of a Differential Equation

An ordinary differential equation is obtained when we eliminate arbitrary constants (also called parameters) from a given relation involving the variables; the order of the equation being equal to the number of the constants to be eliminated. A partial differential equation is obtained when we eliminate arbitrary constants or functions from a given relation involving the variables/functions. We consider here only ordinary differential equations and hence by a differential equation we mean an ordinary differential equation.

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Formation of a Differential Equation

1-5

Differential equations occur in physical, biological, chemical, engineering and geometrical problems, and also in problems relating to economics, ecology, population studies, medicine, etc.

φ ( x, y, c1 , c2 ,…, cn ) = 0

Let

(1.1)

be a relation involving variables x and y and n arbitrary constants c1 , c2 ,… , cn . If we differentiate Eq. (1.1) n times with respect to ‘x’ and eliminate the n arbitrary constants between the (n + 1) relations, we get an ordinary differential equation of the form F ( x, y, y ′, y ′′, , y ( n ) ) = 0

(1.2)

where y′ = y ( n)

dy d2 y , y ′′ = 2 , and dx dx dn y = n. dx

Example 1.2.1 Eliminate a and b from y = ax 2 + bx and form the differential equation. Solution y = ax 2 + bx

(1.3)

Differentiating twice with respect to ‘x’ we have y ′ = 2ax + b

(1.4)

y ′′ = 2a

(1.5)

1 y ′′ 2 (1.4) ⇒ b = y ′ − 2ax = y ′ − xy ′′ 1 (1.3) ⇒ y = y ′′ ⋅ x 2 + ( y ′ − xy ′′ ) x 2 1 = xy ′ − x 2 y ′′ 2 x 2 y ′′ − 2 xy ′ + 2 y = 0. (1.5) ⇒ a =

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1-6

Differential Equations

Example 1.2.2 Form the differential equation by eliminating the constants A and B from y = Ae 2 x + Be −2 x

(1.6)

Solution Differentiating Eq. (1.6) twice with respect to ‘x’ y ′ = 2 Ae 2 x − 2 Be −2 x

(1.7)

y ′′ = 4 Ae + 4 Be

(1.8)

2x

−2 x

From Eqs. (1.6) and (1.8) we obtain y ′′ − 4 y = 0 which is the required differential equation. Example 1.2.3 Obtain the differential equation by eliminating λ from x2 y2 + =1 a2 + λ b2 + λ

(1.9)

Solution Differentiating Eq. (1.9) with respect to ‘x’ once 2x 2 yy ′ + =0 a2 + λ b2 + λ x yy ′ x + yy ′ ⇒ 2 = = 2 2 a + λ − (b + λ ) a − b2

(1.10)

⎛ Sum of numerators ⎞ ⎜⎝ = Sum of denominators ⎟⎠ ⇒ and Substituting in Eq. (1.9)

1 x + yy ′ = a + λ x ( a2 − b2 ) 2

x + yy ′ 1 = b 2 + λ − yy ′( a 2 − b 2 )

(1.11)

y ( x + yy ′ ) x( x + yy ′ ) + 2 2 ( a − b ) − y ′( a 2 − b 2 ) ( x + yy ′ ) ( xy ′ − y ) = y ′( a 2 − b 2 )

1=

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Formation of a Differential Equation

or

1-7

( x + yy ′ ) ( xy ′ − y ) = ( a 2 − b 2 ) y ′

which is the required differential equation. Example 1.2.4 Determine the differential equation by eliminating the constants A and B from y = A sin x + B cos 2 x

(1.12)

Solution Differentiating Eq. (1.12) twice with respect to ‘x’ y ′ = A cos x − 2 B sin 2 x

(1.13)

y ′′ = − A sin x − 4 B cos 2 x

(1.14)

From Eqs. (1.12) and (1.14), we obtain y ′′ + 4 y = 3 A sin x

(1.15)

y ′′ + y = −3 B cos 2 x

(1.16)

Substituting the values of A and B in Eq. (1.13) ⎛ y ′′ + 4 y ⎞ ⎛ y ′′ + y ⎞ + 2sin 2 x ⎜ y ′ = cos x ⎜ ⎟ ⎝ 3sin x ⎠ ⎝ 3cos 2 x ⎟⎠ ⇒ 3 y ′ = y ′′(cot x + 2 tan 2 x ) + y (4 cot x + 2 tan 2 x )

(1.17)

which is the required differential equation. Example 1.2.5 Find the differential equation of the family of curves y = Ae 2 x + Be −2 x for different values of A and B Solution y = Ae 2 x + Be −2 x

(1.18)

Differentiating (1.18) w.r.t. x

dy = 2 Ae 2 x − 2 Be-2 x dx Differentiating (1.19) w.r.t. x d2 y = 4 Ae 2 x + 4 Be -2 x dx 2

Ch01.indd 7

(1.19)

(1.20)

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1-8

Differential Equations

Eliminating A and B from (1.18) and (1.20) we obtain the differential equation

d2 y =4y dx 2

(1.21)

Example 1.2.6 Find the differential equation by eliminating a, b, c from y = ae 2 x + be −3 x + ce x Solution y = ae 2 x + be −3 x + ce x

(1.22)

Differentiating (1.22) w.r.t. x dy = 2ae 2 x − 3be −3 x + ce x dx Differentiating (1.23) w.r.t. x d2 y = 4 ae 2 x + 9be −3 x + ce x d2x Differentiating (1.24) w.r.t. x d3 y = 8ae 2 x − 27be −3 x + ce x dx 3

(1.23)

(1.24)

(1.25)

Now dy − 6 y = 14 ae 2 x − 21be −3 x + 7ce x − 6ae 2 x − 6be −3 x − 6ce x dx d3 y = 8ae 2 x – 27be –3 x + ce x = dx 3 ∴ The required differential equation is 7

d3 y dy − 7 + 6y = 0 3 dx dx

(1.26)

Example 1.2.7 By eliminating the constants a and b from xy = ae x + be – x + x 2 obtain the differential equation

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Formation of a Differential Equation

1-9

Solution xy = ae x + be – x + x 2

(1.27)

Differentiating (1.27) w.r.t. x

xy1 + y = ae x – be – x + 2 x

(1.28)

Differentiating (1.28) w.r.t. x

xy2 + 2 y1 = ae x + be – x + 2

(1.29)

Eliminating a and b from (1.27) and (1.29)

xy2 + 2 y1 – xy + x 2 – 2 = 0

(1.30)

Example 1.2.8 Show that Ax 2 + By 2 = 1 is a solution of

⎡ d 2 y ⎛ dy ⎞ 2 ⎤ dy x ⎢y 2 + ⎜ ⎟ ⎥ − y =0 ⎝ dx ⎠ ⎦⎥ dx ⎣⎢ dx Solution

Ax 2 + By 2 = 1

(1.31)

Differentiating (1.31) w.r.t. x

2 Ax + 2 Byy1 = 0

(1.32)

Differentiating (1.32) w.r.t. x

(

)

2 A + 2 B yy2 + y12 = 0

(1.33)

Eliminating A and B from (1.32) and (1.33)

(

)

x yy2 + y12 – yy1 = 0 where y1 =

dy

d2 y = and y 2 dx dx 2

Example 1.2.9 Find the differential equation of the family of curves 2 y = c ( x – c ) where c is an arbitrary constant.

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1-10

Differential Equations

Solution

y = c ( x – c)

2

(1.34)

dy = 2c ( x − c) dx

Differentiating (1.34) w.r.t. x

(1.35)

Squaring (1.35) and dividing it by (1.34) 2

⎛ dy ⎞ ⎜⎝ ⎟⎠ dx = 4c or y

1 c= 4y

⎛ dy ⎞ ⎜⎝ ⎟⎠ dx

2

(1.36)

Substituting the value of c in (1.35)

⎛ dy ⎞ ⎜⎝ ⎟⎠ dy dx =2 dx 4y

2 2 3 ⎡ 1 ⎛ dy ⎞ ⎤ dy ⎛ dy ⎞ 2 ⎢x − ⎜ ⎟ ⎥ or 8 y = 4 xy − ⎜⎝ ⎟⎠ (1.37) 4 y ⎝ dx ⎠ ⎥⎦ dx dx ⎢⎣

which is the required differential equation. Example 1.2.10 Find the differential equation of the family of ellipses in the standard form. Solution The equation of the family of ellipses in the standard form is

x2 y2 + =1 a2 b2 where a and b are parameters. Differentiating (1.38) w.r.t. x 2 x 2 y dy y dy b2 + =0⇒ =− 2 a2 b2 dx x dx a Differentiating (1.39) w.r.t. x 2

y d y ⋅ + x dx 2

x

dy − y.1 dy dx ⋅ =0 2 x dx

(1.38)

(1.39)

(1.40)

2

d2 y dy ⎛ dy ⎞ + x⎜ ⎟ − y =0 2 ⎝ dx ⎠ dx dx which is the required differential equation. ⇒ xy

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Formation of a Differential Equation

1-11

Example 1.2.11 Obtain the differential equation of the family of circles with centres on the x-axis and passing through the origin. Solution The equation of the family of circles passing through the origin and with centres on the x-axis is

x 2 + y 2 + 2ax = 0

(1.41)

where ‘a’ is a parameter Differentiating (1.41) w.r.t. x

2x + 2 y

dy dy + 2a = 0 ⇒ a = − x − y dx dx

(1.42)

Putting the value of a in (1.41)

dy ⎞ dy ⎛ x 2 + y 2 − 2 x ⎜ x + y ⎟ = 0 ⇒ y 2 − x 2 − 2 xy = 0 (1.43) ⎝ ⎠ dx dx which is the required differential equation. Example 1.2.12 Find the differential equation of the family of parabolas having vertex at the origin and focii on the y-axis. Solution The equation of the family of parabolas with vertex at the origin and focii on the y-axis is

x 2 = 4 ay

(1.44)

where a is a parameter Differentiating (1.44) w.r.t. x

2 x = 4a

dy 2x ⇒ 4a = dx ⎛ dy ⎞ ⎜⎝ ⎟⎠ dx

(1.45)

Eliminating a from (1.44) and (1.45)

⎛ ⎞ ⎜ 2x ⎟ x2 = ⎜ ⎟y ⎜ ⎛⎜ dy ⎞⎟ ⎟ ⎝ ⎝ dx ⎠ ⎠

or

x

dy = 2y dx

(1.46)

which is the required differential equation.

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1-12

Differential Equations

1.2.2

Solution of a Differential Equation A relation φ ( x, y ) = 0 defining a function y = f ( x ) in some

interval I, which has derivatives f ′, f ′′, , f ( n ) such that F ( x, f , f ′, f ′′, , f ( n ) ) = 0, i.e., satisfying the differential equation, is called a solution of the differential equation: F ( x, y, y ′, y ′′, , f ( n ) ) = 0. General (or complete) solution A relation φ ( x, y, c1 , c2 ,, cn ) = 0 containing n independent arbitrary constants ci which is a solution of the differential equation F ( x, y, y ′, y ′′, , y ( n ) ) = 0 is called the general (complete) solution of the differential equation. Particular solution Any solution obtained from the general solution of a differential equation, by giving particular values to the arbitrary constants in it, is called a particular solution of the differential equation. Singular solution A solution φ ( x, y ) = 0 of the differential equation F ( x, y, y ′, y ′′, , y ( n ) ) = 0 which is neither a general solution nor a particular solution of it, is called a singular solution of the differential equation. Only some equations have singular solutions. In the context of differential equations, solution and ‘integral’ have the same meaning; and the general solution is sometimes called the primitive. Example 1.2.13 y = Ae 3 x + Be −3 x , where A and B are arbitrary constants, is the general solution while y = −2e 3 x

and

y = 10e −3 x

are particular solutions of the second order and first degree differential equation

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Formation of a Differential Equation

1-13

d2 y − 9y = 0 dx 2 It has no singular solution. Example 1.2.14 y = ( x + a)2 is the general solution; y = x 2 is a particular solution; and y = 0 is a singular solution of the first order and second degree differential equation 2

⎛ dy ⎞ ⎜⎝ ⎟⎠ − 4 y = 0. dx Note that the singular solution y = 0 cannot be obtained from the general solution for any value of the constant ‘a’. Formation of a differential equation is straight-forward while the solution of a differential equation is not. So, we have to classify them and find methods of solution. EXERCISE 1.1 Form the differential equation in each of the following cases by eliminating the parameters mentioned against each. 1. y = ax + bx 2

( a, b )

Ans: 2 y + x 2 y ′′ = 2 xy ′ 2. x = A cos( pt + B )

( A, B)

Ans: x ′′ + p 2 x = 0 (′ denotes differentiation with respect to ‘t’) 3. y = mx +

a m

Ans: y = xy ′ +

( m) a y′

4. y = ax 2 + bx + c

( a, b, c )

Ans: y ′′′ = 0

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1-14

Differential Equations

5. ( x − h)2 + ( y − k )2 = r 2

( h, k )

dy d2 y⎞ ⎛ Ans: (1 + y12 )3 = r 2 y22 , ⎜ y1 = , y2 = 2 ⎟ ⎝ dx dx ⎠ 6. y = e x ( A cos x + B sin x )

( A, B )

Ans: y2 − 2 y1 + 2 y = 0 7. Find the differential equation for the family of circles with their centres on the x-axis. (Hint: x 2 + y 2 + 2 gx + c = 0 g, c parameters) Ans: 1 + y12 + yy2 = 0 8. Form the differential equation for the family of circles, touching the x-axis at (0, 0). (Hint: x 2 + y 2 − 2 fy = 0, f parameter) Ans: ( y 2 − x 2 ) y ′ + 2 xy = 0 9. Form the differential equation of all parabolas each having its latus-return = 4a and its axis parallel to the x-axis. (Hint: ( y − k )2 = 4 a( x − h); h, k parameters) Ans: 2ay2 + y13 = 0 10. Find the differential equation by eliminating c from y = cx + x 3. dy − y − 2 x3 = 0 dx 11. Show that y = ax + b e x , where a and b are arbitrary constants, dy ⎞ ⎛ is a solution of ( x – 1) y ″ – xy ′ + y = 0 ⎜ y ′ = ⎟ . ⎝ dx ⎠ 12. Show that the differential equation obtained from y = Ae ax cos bx + Be ax sin bx where A and B are arbitrary constants is y ″ – 2ay ′ + ( a 2 + b 2 ) y = 0. Ans: x

13. Find the differential equation of the family of circles in the xy plane. (Hint: The equation of the circles is x 2 + y 2 + 2 gx + 2 fy + c = 0 ) Ans: (1 + y12 ) y3 − 3 y1 ( y22 ) = 0

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Formation of a Differential Equation

1-15

14. Find the differential equation of the family of curves x2 y2 + = 1 where l is a parameter. a2 a2 + λ Ans: ( x 2 – a 2 ) y ′ = xy 15. Find the differential equation of all the ellipses with their centres at the origin. Ans: xyy ″ + x ( y ′ )2 – yy ′ = 0 16. Show that the differential equation of all parabolas having their axes of symmetry coincident with the x-axis is 2 d 2 y ⎛ dy ⎞ y 2 + ⎜ ⎟ = 0. ⎝ dx ⎠ dx

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2 Differential Equations of First Order and First Degree 2.1

FIRST ORDER AND FIRST DEGREE DIFFERENTIAL EQUATIONS

The simplest type of the ordinary differential equations is that of the first order and first degree and is of the form dy ⎞ ⎛ F ⎜ x, y, ⎟ = 0 ⎝ dx ⎠

dy we can write Eq. (2.1) as dx dy = f ( x, y ) dx and if we write

(2.1)

Solving for

f ( x, y ) = −

(2.2)

M ( x, y ) N ( x, y )

then Eq. (2.2) can be put in the form Mdx + Ndy = 0

Ch02.indd 1

(2.3)

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2-2

Differential Equations

where M and N are functions of x and y. In order to study the methods of solution, we classify the first order and first degree equations as follows: 1. 2. 3. 4. 5. 6. 7. 8.

Variables separable Reducible to variables separable Homogeneous equation Non-homogeneous but reducible to homogeneous type Exact equation Inexact equation but rendered exact using integrating factors First order linear equation. Reducible to linear equation i. Bernoulli’s equation ii. Others

Now we take up the above types of equations one by one and discuss the methods of solution.

2.1.1 Variable Separable Equation If the first order and first degree differential equation is of the form f1 ( x ) g2 ( y )dx + f 2 ( x ) g1 ( y )dy = 0 then it can be written as M ( x )dx + N ( y )dy = 0 where M ( x ) = ( f1 /f 2 )( x ), is a pure function of x; and N ( y ) = ( g1 /g2 )( y ), a pure function of y, so that the equation can be readily integrated and the solutions obtained. Example 2.1.1 Solve e −4 y log x dx + x cos3 y dy = 0. Solution If we multiply the equation by 1 4y cos ydy = 0. Integrating, we get x log xdx + e

1 x

e4 y

it becomes

2 (log x )2 + e 4 y (4 cos3 y + 3sin 3 y ) = c 5 where c is an arbitrary constant. Example 2.1.2 Solve

Ch02.indd 2

„

dy x(2log x + 1) = . dx sin y + y cos y

12/9/2011 11:36:42 AM

Differential Equations of First Order and First Degree

2-3

x(2log x + 1)dx − Solution Separating the variables, we have 2 − (sin y + y cos y )dy = 0. Integrating, we get x log x = y sin y + c. „ Example 2.1.3 Solve

dy ( x 2 − 1)( y + 2) = . dx ( y 2 − 1)( x + 2)

Solution Separating the variables, x2 − 1 y2 − 1 dx = dy x+2 y+2 ⇒

( x 2 − 22 ) + 3 ( y 2 − 22 ) + 3 dx = dy x+2 y+2

⎛ 3 ⎞ 3 ⎞ ⎛ ⇒⎜x −2+ dx = ⎜ y − 2 + dy ⎟ ⎝ x + 2⎠ y + 2 ⎟⎠ ⎝ Integrating, we get x2 − 2 x + 3log x + 2 2 y2 = − 2 y + 3log y + 2 + c1 2 x+2 ⇒ x 2 − y 2 − 4( x − y ) + 6log = 2c1 y+2 x+2 ⇒ ( x − y )( x + y − 4) + 6log =c y+2 where c = 2c1 . Example 2.1.4 Solve

„ dy = ( x + y + 2)2. dx

Solution Put x + y + 2 = u ⇒ or

Ch02.indd 3

dy du = − 1 = u2 dx dx du = dx 2 u +1

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2-4

Differential Equations

Integrating, we get

Example 2.1.5 Solve

x+c=

1 −1 tan ( x + y + 2) 2

(1 +

x 2 ) dy = (1 + y 2 ) dx.

„

Solution

(1 + x 2 ) dy = (1 + y 2 ) dx

(2.4)

Separating the variables dy dx = 2 1+ y 1 + x2 Integrating (2.5) we get tan –1 y = tan –1 x = tan –1 c ⇒ tan –1 y – tan –1 x = tan –1 c ⇒ tan −1

(2.6)

y–x = tan –1 c 1 + yx

y–x = c or ( y – x ) = c (1 + yx ) 1 + yx

or

(2.5)

(2.7) „

Example 2.1.6 Solve

(1 + e x ) y dy = ( y + 1) e x dx.

Solution

(1 + e x ) y dy = ( y + 1) e x dx

(2.8)

Separating the variables we get y ex dy = dx 1 + ex y +1 ⎛ 1 ⎞ ex ⇒ ⎜1 − dy = dx y + 1⎟⎠ 1 + ex ⎝

(2.9)

Integrating (2.9) we get y – log ( y + 1) = log (1 + e x ) + log c

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Differential Equations of First Order and First Degree

2-5

y = log c ( y + 1) (1 + e x )

( y + 1) (1 + e x ) = Ae y

or

1⎞ ⎛ ⎜⎝ A = ⎟⎠ c

(2.10) „

Example 2.1.7 Solve 3e x (1 + tan y ) dx + (1 – e x ) sec2 y dy = 0. Solution 3e x (1 + tan y ) dx + (1 – e x ) sec2 y dy = 0

(2.11)

Separating the variables we get sec2 y dy 3e x dx + =0 x 1− e 1 + tan y

(2.12)

Integrating (2.12) we get –3log (1 – e x ) + log (1 + tan y ) = log c

(2.13)

Dropping logs we have 1 + tan y =c (1 − e x )3

or

1 + tan y = c(1 − e x )3

(2.14) „

Example 2.1.8 Solve

dy = sin( x + y ) + cos( x + y ). dx

Solution dy = sin( x + y ) + cos( x + y ) dx

(2.15)

dy du dy du = or = − 1. dx dx dx dx The given equation becomes

Let x + y = u so that 1 +

du du − 1 = sin u + cos u ⇒ = 1 + sin u + cos u dx dx

Ch02.indd 5

(2.16)

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2-6

Differential Equations

Separating the variables du = dx or 1 + cos u + sin u

2cos

2 u 2

du = dx + 2sin u2 cos u2

(2.17)

This can be written in the form sec2 u2 du = dx 1 + tan u2

(2.18)

u⎞ ⎛ log ⎜1 + tan ⎟ = x + const ⎝ 2⎠

(2.19)

1 2

Integrating we get

Dropping logs we get 1 + tan

u = Ae x 2

⎛ x + y⎞ 1 + tan ⎜ = Ae x ⎝ 2 ⎟⎠

or

Example 2.1.9 Solve (1 + x 2 ) xy

(2.20) „

dy = (1 + y 2 )(1 + x + x 2 ). dx

Solution (1 + x 2 ) xy

dy = (1 + y 2 )(1 + x + x 2 ) dx

(2.21)

Separating the variables y dy 1 + x + x 2 = dx 1 + y 2 x(1 + x 2 ) 1 ⎞ ⎛1 =⎜ + dx ⎝ x 1 + x 2 ⎟⎠

(2.22)

Integrating (2.22) we have 1 log (1 + y 2 ) = log x + tan –1 x + const. 2

Ch02.indd 6

(2.23)

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Differential Equations of First Order and First Degree

2-7

Dropping logs

or

1 + y2 −1 = Ae tan x x

(2.24)

1 + y 2 = Axe tan

(2.25)

−1 x

„ EXERCISE 2.1 Solve: 1.

dy y + b = . dx x + a Ans: x + a = c( y + b)

2. e x tan ydx + (1 − e x )sec2 ydy = 0. Ans: tan y = c(1 − e x ) 3. x 1 + y 2 dx + y 1 − x 2 dy = 0. Ans: 1 + y 2 − 1 − x 2 = c 4.

1 − x 2 sin −1 xdy + ydx = 0. Ans: y sin −1 x = c

5.

dy 1 − y2 + = 0. dx 1 − x2

Ans: sin −1 x + sin −1 y = c 6. x 1 + y 2 dx + y 1 + x 2 dy = 0. Ans: 1 + x 2 + 1 + y 2 = c 7. ydx − xdy + 3 x 2 y 2 e x dx = 0. 2

Ans: ( x y ) + e x = c 3

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2-8

Differential Equations

8. ( y 2 − x 2 ) y ′ + 2 xy = 0. Ans: x 4 + 2e 2 x = c + 4e y 1 + x 2 1 + y 2 dx + dy = 0.

9.

x 1 1 + x 2 + sinh −1 x + sinh −1 y = c 2 2 dy 10. 2 xe x2 + y2 = (1 + 2 y )e − y . dx 2 2 Ans: e x + e y + y = c Ans:

dy = a2 . dx a x− y−a Ans: y = log +c 2 x− y+a dy = cos( x + y ). 12. dx

11. ( x − y )2

Ans: tan[( x + y ) 2] = x + c 13.

dy x(2log x + 1) = . dx sin y + y cos y Ans: y sin y = x 2 log x + c dy + y( y − 1) = 0. dx Ans: ( x − 1)( y − 1) = cxy

14. x( x − 1)

15. ( x 2 + 1)

dy + y2 + 1 = 0 dx

y(0) = 1.

[JNTU 2001]

Ans: x + y + xy = 1 dy + y + 4 = 0. dx Ans: x( y + 4) = c

16. x

17.

Ch02.indd 8

[JNTU 2001]

dy x + x 2 = . dx y − y 2 y2 y3 x2 x3 Ans: − = + + c. 2 3 2 3

12/9/2011 11:36:44 AM

Differential Equations of First Order and First Degree

18. y − x

2-9

dy dy ⎞ ⎛ = 3 ⎜1 + x 2 ⎟ . ⎝ dx dx ⎠

Ans: ( y – 3)(3 x + 1) = cx 19. ( e y + 1) cos x dx + e y sin x dy = 0. Ans: (1 + e y ) sin x = c 20. y 2 cos x – 2 x e y dy = 0. 1

1

Ans: sin x + e y = c 21. 3e x cos 2 y dx + (1 + e x ) cot y dy = 0. Ans: tan y = c (1 + e x )

3

22. cos ( x + y ) dy = dx. ⎛ x + y⎞ Ans: y = c + tan ⎜ ⎝ 2 ⎟⎠ dy 23. + 1 = e x + y. dx

Ans: ( x + c ) e x + y + 1 = 0 ⎛ x + y − a ⎞ dy ⎛ x + y + a ⎞ = 24. ⎜ . ⎝ x + y − b ⎟⎠ dx ⎜⎝ x + y + b ⎟⎠ Ans: ( x + y ) +

1 (b − a) log ⎡⎣( x + y )2 − ab⎤⎦ = 2 x + c 2

2.1.2 Homogeneous Equations A polynomial f (x, y) is called a homogeneous function of degree n if f (tx, ty ) = t n f ( x, y ) for some t ∈R or equivalently, f ( x, y ) = x n f (1, y x ). Example 2.1.10 f ( x, y ) = 2 x + 3 y is a homogeneous function of degree 1 since y⎞ y ⎛ f ( x, y ) = x ⎜ 2 + 3 ⎟ = xf (1, ). ⎝ ⎠ x x

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2-10

Differential Equations

Example 2.1.11 g ( x, y ) = 3 x 2 − 4 xy + 7 y 2 is a homogeneous function of degree 2 since 2 ⎡ ⎛ y⎞ ⎛ y⎞ ⎤ g ( x, y ) = x ⎢3 − 4 ⎜ ⎟ + 7 ⎜ ⎟ ⎥ = x 2 f (1, y x ). ⎝ x⎠ ⎝ x ⎠ ⎥⎦ ⎢⎣ 2

A first order and first degree differential equation dy = f ( x, y ) dx

(2.26)

is called a homogeneous type if f ( x, y ) is a rational function which is homogeneous and of degree zero. That is, f ( x, y ) is the quotient of two polynomials each of the same degree n. If we put y = vx (or x = vy ) Eq. (2.26) transforms to v+x

⎛ dv dv 1 ⎞ = f (1, v ) or ⎜ v + y = dx dy f ( v,1) ⎟⎠ ⎝

where the variables are separable. Separating the variables and integrating, we get the general solution as dv ⎛ ⎞ x = c exp ∫ ⎜ f (1, v ) − v ⎟ ⎜ f ( v,1) dv ⎟ ⎜ or y = c exp ∫ ⎟ 1 − vf ( v,1) ⎠ ⎝

(2.27)

where v = y x (or v = x y ) and c is an arbitrary constant. We extend the meaning of ‘homogeneity’ to include differential equations with functions of the following type since they are solvable by the above method. ⎛ x2 + y2 ⎞ x y e + sin( y x ). Example 2.1.12 f ( x, y ) = ⎜ ⎝ xy ⎟⎠ Example 2.1.13 f ( x, y ) = sin

x2 + y2 + tan −1 ( y x ). x2 − y2

Example 2.1.14 f ( x, y ) = log( x 2 − 2 xy + y 2 ) / ( x 2 + y 2 ).

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Differential Equations of First Order and First Degree

Example 2.1.15 Solve

2-11

dy y2 = . dx xy + x 2

Solution Put y = vx ⇒

dy dv =v+x dx dx

The equation transforms to v+x

dv v2 dy v 2 − v 2 − v = 2 ⇒x = dx v + 1 dx v +1

Separating the variables

dx v + 1 + dv = 0 x v

Integrating, log x + log v + v = constant Dropping logarithms we may put the solution as y

ye x = c „

where c is an arbitrary constant. Example 2.1.16 Solve xdy − ydx = x 2 + y 2 dx. Solution Put y = vx ⇒ dy = vdx + xdv The equation becomes x( vdx + xdv ) − vxdx = x 1 + v 2 dx Separating the variables dx dv = x 1 + v2 Integrating log x − log( v + 1 + v 2 ) = Constant y Dropping logarithms, replacing v by x and simplifying we get y + x 2 + y 2 = cx 2 where c is an arbitrary constant.

Ch02.indd 11

„

12/9/2011 11:36:45 AM

2-12

Differential Equations

Example 2.1.17 Solve Solution Put

(1 + e ) dx + e x y

x y

⎛ x⎞ ⎜⎝1 − y ⎟⎠ dy = 0.

x = vy ⇒

dx dv =v+ y dy dy

The equation transforms to dv e v (1 − v ) =− dy 1 + ev v dv e (1 − v ) ⇒y =− −v dy 1 + ev − e v (1 − v ) − v(1 + e v ) = 1 + ev

v+ y

Separating the variables we have dy 1 + e v dv = 0 + y v + ev ⇒ y( v + e v ) = c,on integration x ⇒ x + ye y = c

„

where c is an arbitrary constant. Example 2.1.18 Solve ( ydx + xdy ) x cos Solution ( ydx + xdy ) x cos

y y = ( xdy − ydx ) y sin . x x

y y = ( xdy − ydx ) y sin x x

(2.28)

Eq. (2.28) can be written as y y⎤ y y⎤ ⎡ ⎡ 2 2 ⎢⎣ xy cos x + y sin x ⎥⎦ dx = ⎢⎣ xy sin x − x cos x ⎥⎦ dy dy xy cos xy + y 2 sin xy = dx xy sin xy − x 2 cos xy

Ch02.indd 12

(2.29)

12/9/2011 11:36:45 AM

Differential Equations of First Order and First Degree

2-13

y This is a homogeneous equation. Putting = ν or y = ν x in x (2.29) we get dν ν cos ν + ν 2 sin ν x = −ν dx ν sin ν − cos ν dν 2ν cos ν = dx ν sin ν − cos ν Separating the variables, we have dx ν sin ν − cos ν 2 = dν x ν cos ν Integrating we have 2log x = – log cos ν – log ν + log c x

⇒ log x 2 =

log c y ⇒ xy cos = c x (ν cos ν )

(2.30)

(2.31)

(2.32)

x x ⎛ ⎞ x Example 2.1.19 Solve ⎜1 + e y ⎟ dx + e y (1 − ) dy = 0. y ⎝ ⎠ x x ⎛ ⎞ x Solution ⎜1 + e y ⎟ dx + e y (1 − ) dy = 0. y ⎝ ⎠

(2.33)

Eq. (2.33) can be written as x

(

)

x dx e y y − 1 = x dy 1+ e y

Putting x = ν y we get y

dν eν (ν − 1) eν + ν = − ν = − 1 + eν 1 + eν dy

Separating the variables and integrating

∫

dy 1 + eν = −∫ dν + log c ν + eν y

y(ν + eν ) = c

x ⎞ ⎛x y⎜ y + ey ⎟ = c ⎝ ⎠

or x

or

Ch02.indd 13

x + ye y = c

„

12/9/2011 11:36:46 AM

2-14

Differential Equations

EXERCISE 2.2 Solve: 1.

y 2 = ( xy − x 2 )

dy . dx

y

Ans: y = ce x

2. xdy − ydx = x 2 + y 2 dx. Ans: cx 2 = x 2 + y 2 + y 3.

( x+ y ) dy y = +e x . dx x

Ans: e

( x+ y ) x

log x = c

4. 2 xydy = ( x 2 + y 2 ).

[JNTU 1999]

Ans: cx = ( x − y ) 5. x 2 ydx = ( x 3 + y 3 )dy. 2

2

x

Ans: y = ce y 6. ( x 2 − y 2 )dx = xydy. Ans: x 2 ( x 2 − 2 y 2 ) = c y y dy y 7. x tan + x sec2 ⋅ = y sec2 . x x dx x y Ans: x tan = c x x x dy . 8. ye y = xe y + y dx

(

)

x

Ans: y = ce y y dy y = y sin + x. x dx x y y⎞ ⎛ Ans: x 2 ⎜ sin + tan ⎟ = c ⎝ x x⎠ 3 2 dy y + 3 x y 10. dx = x 3 + 3 xy 2 . 9. x sin

Ans: x10 = cy 6 ( x 2 − y 2 )

Ch02.indd 14

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Differential Equations of First Order and First Degree

2-15

x x ⎛ x⎞ 11. 1 + 2e y dx + 2e y ⎜1 − ⎟ dy = 0. y⎠ ⎝

(

)

Ans: xe cos( y x ) = c dy y 2 12. x + = y. dx x ⎛ x⎞ y Ans: sin −1 ⎜ ⎟ = log c ⎝ y⎠ ⎡ ⎛ x⎞ ⎛ x⎞⎤ 13. xy log ⎜ ⎟ dx + ⎢ y 2 − x 2 log ⎜ ⎟ ⎥ dy = 0. ⎝ y⎠ ⎝ y⎠ ⎦ ⎣ ⎡ ⎛ y⎞ ⎤ ⎢2log ⎜⎝ x ⎟⎠ + 1⎥ = log cy ⎣ ⎦ dy y = . 14. dx x + xy Ans:

x2 4 y2

Ans: y = ce

2

[JNTU 1995]

x y

15. ( y 2 − 2 xy )dx = ( x 2 − 2 xy )dy. Ans: xy( x − y ) = c dy y y 16. = + tan . dx x x

[JNTU 1995]

y

Ans: x = c sin x . 17. x sin

y dy y = y sin − x. x dx x

⎛

y x

⎞

Ans: x = c exp ⎜ cos ⎟ ⎝ ⎠ 18.

dy = dx

y x + ye

−2 x y

.

⎛ 2x ⎞ ⎝ y ⎟⎠

Ans: 2 ( c + log y ) = exp ⎜ 19.

Ch02.indd 15

( x – y log y + y log x ) dx + x (log y – log x ) dy = 0. Ans: ( x – y ) log x + y log y = cx + y

12/9/2011 11:36:46 AM

2-16

Differential Equations

20. x 2 dy = ( x 2 + xy + y 2 ) dx. Ans: sinh −1

y = log x + c x

2.1.3 Non-homogeneous Equations The first order and first degree differential equation of the type dy ax + by + c = dx lx + my + n

(2.34)

where a, b, c, l, m, n are given real constants is called a nonhomogeneous differential equation. If c = n = 0 then Eq. (2.34) reduces to a homogeneous equation dY aX + bY whose solution we have discussed above. The new = dX lX + mY method of solution (the method of proportions) used for solving nonhomogeneous equations (case (b)) below may be used for solving the homogeneous equation of this type. But other forms under homogeneous type have to be solved only by putting y = vx or x = vy. We assume that at least one of c and n is not zero. In general, the following two cases are discussed. a b c = ≠ . In this case, Eq. (2.34) can be solved by putting l m n u = ax + by or u = lx + my and separating the variables. Case (A)

a b ≠ . In this case, Eq. (2.34) is solved by shifting the origin l m to a point (h, k) by the substitution x = X + h, y = Y + k and choosing (h, k) such that ah + bk + c = 0; lh + mk + n = 0 which is possible since

Case (B)

a b ⎛ bn − cm cl − an ⎞ = am − bl ≠ 0. Indeed, we have ( h, k ) = ⎜ , . ⎝ am − bl am − bl ⎟⎠ l m dY aX + bY = Eq. (2.34) now reduces to the homogeneous type dX lX + mY which can be solved by putting Y = vX or X = vY and separating the variables. This method is lengthy and tedious.

Ch02.indd 16

12/9/2011 11:36:47 AM

Differential Equations of First Order and First Degree

2-17

By dividing the problem into various cases, we can apply simpler and easier methods to solve the problem. Finally, the technique described below, based on the application of the property of proportion makes solution of the problem simpler and easier in the above two cases, as well. In this section, C is used to denote an arbitrary constant instead of c. a b c dy = = = λ (say). Eq. (2.34) reduces to = λ whose l m n dx general solution is y = λ x + c.

Case (1)

Case (2) (i) b = l = 0 or (ii) a = m = 0. In either case, the variables are separable and hence Eq. (2.34) is integrable. Case (3) c ≠ 0 or n ≠ 0 or both c ≠ 0 and n ≠ 0. (i) a = b = 0 (c ≠ 0)

(ii) l = m = 0 ( n ≠ 0)

In case (3)(i) Eq. (2.34) reduces to dy c = dx lx + my + n

(2.35)

which can be solved by putting u = lx + my and in case (3)(ii), Eq. (2.34) reduces to dy ax + by + c = dx n which can be solved by putting u = ax + by.

(2.36)

Alternative method (Method of proportions) By the property of proportions, we can write Eq. (2.35) in case (3) as dx dy = lx + my + n c using (l , m) as multipliers, each ratio =

Ch02.indd 17

dy d (lx + my ) = c l (lx + my ) + ln + mc

12/9/2011 11:36:48 AM

2-18

Differential Equations

Integrating, the last two ratios give 1 1 y = log(lx + my + mc l ) + const. c l or, dropping logs we get (lx + my + mc l )c = Ce ly Now, by the property of proportions, we can write Eq. (2.36) in case (3)(ii) as dx dy = n ax + by + c Using ( a, b) as multipliers, each ratio =

d ( ax + by ) dx = n b ( ax + by + c) + an

Integrating, the last two ratios give 1 1 x = log( ax + by + an b) + const. n b or dropping logs we get ( ax + by + an b) n = Ce bx Case (4) b + l = 0. Separating the variables and grouping the terms, we can write Eq. (2.34) as ( ax + c) dx + bd ( xy ) − ( my + n) dy = 0 ∵

l = − b and ydx + xdy = d ( xy )

Integrating, we obtain the general solution as ax 2 + 2bxy − my 2 + 2cx − 2ny + C = 0 This is an exact equation. We will study exact equations in detail is Section 2.1.4.

Ch02.indd 18

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Differential Equations of First Order and First Degree

2-19

Case (5) a = m, b = l ≠ 0. dy ax + by + c = dx bx + ay + n

Eq. (2.34) becomes

(2.37)

This falls under case (A) if a = b and case (B) if a ≠ b. But it can be solved simply by the method of proportions as follows. Eq. (2.37) can be written as dy dx = bx + ay + n ax + by + c Sum of numerators Difference of numerators ⎤ ⎡ ⎢⎣ Each ratio = Sum of denominators = Difference of denominators ⎥⎦ =

d ( y + x) d ( y − x) = ( a + b)( y + x ) + n + c ( a − b)( y − x ) + ( n − c)

(2.38)

Integrating we have 1 n + c⎞ 1 n − c⎞ ⎛ ⎛ log ⎜ y + x + log ⎜ y − x + ⎟⎠ = ⎟ + const. ⎝ ⎝ a+b a+b a−b a − b⎠

or, dropping logs we get n + c⎞ ⎛ ⎜⎝ y + x + ⎟ a + b⎠

a−b

n − c⎞ ⎛ =C⎜ y − x+ ⎟ ⎝ a − b⎠

a+b

(2.39)

Case (6) a = − m, b = − l . Eq. (2.34) can be written as dy dx d ( y + x) = = ax + by + c − bx − ay + n (b − a)( y − x ) + c + n d ( y − x) = (b + a)( y + x ) + c − n Cross-multiplying we get [(b + a)( y + x ) + c − n] d ( y + x ) = [(b − a)( y − x ) + c + n] d ( y − x )

Ch02.indd 19

12/9/2011 11:36:49 AM

2-20

Differential Equations

Integrating we have (b + a)( y + x )2 + ( a − b)( y − x )2 + 2(c − n)( y + x ) − 2(c + n)( y − x ) = K Note 2.1.20 This problem is also solvable by grouping terms as in case 4. Finally, we consider case (A) and case (B) mentioned at the beginning. Case (7) b + l ≠ 0. (i)

a b a b ⎛ c⎞ respectively. = = λ ⎜ ≠ ⎟ and (ii) ≠ ⎝ n⎠ l m l m

(i) Alternative method (Method of proportions) Equation (2.34) can be written as dy dx dy = = ax + by + c lx + my + n λ (lx + my ) + c ∵

(2.40)

( a = l λ , b = mλ )

Using ( λ , −1) and (l, m) as multipliers, we obtain from the last equation that each ratio =

d (λ x − y) d (lx + my ) = λn − c (l + λ m)(lx + my ) + nl + cm

(2.41)

Integrating, we get 1 1 (λ x − y) = log[(l + λ m)(lx + my ) + nl + cm] + C (2.42) λn − c l + λm ∵ c ≠ λ n,

l + λ m = b + l ≠ 0.

(ii) Alternative method (Method of multipliers) m ≠ 0; Δ = ( b − l )2 + 4 am ≠ 0 Equation (2.34) can be written as dy dx = ax + by + c lx + my + n

Ch02.indd 20

(2.43)

12/9/2011 11:36:49 AM

Differential Equations of First Order and First Degree

2-21

Using (1, λ ) as multipliers, each ratio =

d ( y + λ x) ( a + λ l ) x + (b + λ m) y + c + λ n

(2.44)

Now, we determine λ such that

λ=

a + λl b + lm

or m λ 2 + (b − l )λ − a = 0

(2.45)

which is a quadratic in l and has two real distinct roots l1, l2 if Δ = (b − l )2 + 4 am > 0. More specifically

λ1 , λ 2 = ⎡⎣ −(b − l ) ± (b − l )2 + 4 am ⎤⎦ 2m ( m ≠ 0). Also, λi =

a + λi l b + λi m

(i = 1,2)

∴ each ratio =

d ( y + λ1 x ) d ( y + λ2 x) 1 = ⋅ b + λ1m y + λ1 x + k1 b + λ 2 m y + λ 2 x + k2 1

⎛ c + λi n ⎜ ki = b + λ m ; ⎝ i

(2.46)

⎞ i = 1,2⎟ ⎠

Integrating and dropping logs, the general solution of equation (2.34) in this case may be written as ( y + λ1 x + k1 )b + λ2 m = C ( y + λ 2 x + k2 )b + λ1m

(2.47)

Note 2.1.21 If Δ < 0 then the roots l1, l2 are complex conjugate numbers of the form a ± ib. The solution (2.47) is valid in this case also. In physical applications we have to separate the real and imaginary parts of (2.47) and take either of them as the solution of the problem. If Δ = 0 or m = 0 the method is not applicable. Then we have to apply the method of case (A), case (B) or case (2)(ii) as suitable.

Ch02.indd 21

12/9/2011 11:36:49 AM

2-22

Differential Equations

Example 2.1.22 Solve

dy 3 x − y + 2 . = dx 6 x − 2 y + 4

Solution Assume that 3 x − y + 2 ≠ 0. Here a = 3, l = 6,

b = −1, m = −2,

c=2 n=4

a b c 1 dy 1 = = = the differential equation reduces to = l m n 2 dx 2 1 (case 1) whose general solution is y = x + C . „ 2

Since

Example 2.1.23 Solve Solution Here a = 2, l = 0,

dy 2 x + 3 = dx 4 y − 5

⎛ ⎜⎝ y ≠

b = 0, m = 4,

c=3 n = −5

5⎞ ⎟. 4⎠

Since b = l = 0, the variables are separable (case 2(i)). Separating the variables, we have (2 x + 3) dx − (4 y − 5) dy = 0 Integrating, we get the general solution as x 2 − 2 y 2 + 3x + 5 y + C = 0 Example 2.1.24 Solve Solution Here a = 0, l = 2,

dy 4 y − 3 = dx 2 x + 5 b = 4, m = 0,

„

5⎞ ⎛ ⎜⎝ x ≠ ⎟⎠ . 2 c = −3 n=5

Since a = m = 0, the variables are separable (case (2)(ii)). Separating the variables, we have dy dy = 4 y − 3 2x + 5 Integrating we have 1 1 log(4 y − 3) = log(2 x + 5) + const. 4 2

Ch02.indd 22

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Differential Equations of First Order and First Degree

2-23

Dropping logs and simplifying we get (4y − 3) = C (2x + 5)2 Example 2.1.25 Solve

„

5 dy (3 x + y ≠ 2). = dx 3 x + y − 2

a = 0, b = 0, c=5 (case (3)(i)) l = 3, m = 1, n=2 Putting u = 3 x + y we can separate the variables and solve.

Solution Here

Alternative method (Method of proportions) The given equation can be written as dx dy = 3x + y − 2 5 Using (3, 1) as multipliers, each ratio =

d (3 x + y ) dy = 9x + 3y − 1 5

Integrating we get 1 1 log(9 x + 3 y − 1) = y + const. 3 5 ⇒

(9 x + 3 y − 1)5 = C e3 y

„

dy x − 2 y + 4 = . dx 3 Solution Here a = 1, b = −2, c = 4 l = 0, m = 0, n = 3 (case (3)(ii)) Putting u = x − 2 y we can separate the variables and solve. Example 2.1.26 Solve

Alternative method (Method of proportions) The given equation can be written as dx dy = 3 x − 2y + 4

Ch02.indd 23

12/9/2011 11:36:50 AM

2-24

Differential Equations

Using (1, –2) as multipliers, each ratio =

d ( x − 2 y) dx = 3 −2( x − 2 y ) − 5

Integrating 1 1 x = − log (2 x − 4 y + 5) + const. 3 2 Dropping logs, the solution can be expressed as (2 x − 4 y + 5)3 e x = C 2

Example 2.1.27 Solve

dy y + x − 2 . = dx y − x − 4

„ (JNTU 1997, 1999)

c = −2 Solution Here a = 1, b = 1, l = −1, m = +1, n = −4

b + l = 0 (case (4))

Separating the variables and grouping the terms we can write the differential equation as ( x − 2) dx + ydx + xdy − ( y − 4)dy = 0 ⇒

(2 x − 4)dx + 2d ( xy ) − (2 y − 8)dy = 0,

on multiplying by 2 and noting that ydx + xdy = d ( xy ). Integrating, we have the general solution as x 2 + 2 xy − y 2 − 4 x + 8 y + C = 0 Example 2.1.28 Solve

Solution Here

a = 2, l = 5,

„

dy 2 x + 5 y + 1 = . dx 5 x + 2 y − 1 b = 5, m = 2,

}

c =1 a b 2 5 = = −21 ≠ 0. n = −1 l m 5 2

Put x = X + h, y = Y + k . The differential equation reduces to dY 2 X + 5Y provided (h, k) are chosen so that 2h + 5k + 1 = 0 and = dX 5 X + 2Y 5h + 2k − 1 = 0.

Ch02.indd 24

12/9/2011 11:36:51 AM

Differential Equations of First Order and First Degree

2-25

By the rule of cross-multiplication h k 1 ⎛ 1 1⎞ = = ⇒ ( h, k ) = ⎜ , − ⎟ ⎝ 3 3⎠ 5( −1) − 2.1 1.5 − 2( −1) 2.2 − 5.5 Put Y = vX ⇒ ⇒

dY dv 2 + 5v =v+ X = dX dX 5 + 2v dv 2 + 5v 2 + 5 v − 5v − 2 v 2 X = −v= 5 + 2v dX 5 + 2v

Separating the variables we have 2⋅

dX 2v + 5 + 2 dv = 0 X v −1

Integrating we get 1 v −1 2log X + log( v 2 − 1) + 5 ⋅ log = const. 2 v +1 5

or

1 ⎛ v − 1⎞ log X 2 ( v 2 − 1) + log ⎜ = const. ⎝ v + 1⎟⎠ 2

Multiplying by 2 and dropping logs we have 5

⎛Y − X ⎞ (Y + X ) (Y − X ) ⎜ = const. ⎝ Y + X ⎟⎠ 2

⇒

2

(Y − X )7 = c1 (Y + X )3

1 1⎞ ⎛ But ( X ,Y ) = ( x − h, y − k ) = ⎜ x − , y + ⎟ so that ⎝ 3 3⎠ 2 Y − X = y − x + ; Y + X = y + x. 3 ∴ The general solution is (3 y − 3 x + 2)7 = C ( y + x )3. Alternative method (Method of proportions) Here a = m = 2, b = l = 5.

Ch02.indd 25

(case (5))

12/9/2011 11:36:51 AM

2-26

Differential Equations

The given differential equation can be written as Sum of numerators dy dx = = 2 x + 5 y + 1 5 x + 2 y − 1 Sum of denominators Difference of numerators = Difference denominators d ( y + x) d ( y − x) = = 7 ( x + y) 3 ( y − x) + 2 Integrating 1 1 log ( x + y ) = log (3 y − 3 x + 2) + const. 7 3 Simplifying we can write the general solution as (3 y − 3 x + 2)7 = C ( y + x )3 Example 2.1.29 Solve

dy 6 x + 5 y − 3 = dx 4 x + y + 2

Solution Here a = 6, l = 4,

b = 5, c = −3 m = 1, n = 2

„

a b 6 5 = = −14 ≠ 0 l m 4 1 b + l ≠ 0. (case (6(ii)))

Put x = X + h, y = Y + k and choose h, k so that 6h + 5k − 3 = 0 and i.e.

4h + k + 2 = 0

h k 1 ⎛ 13 12 ⎞ = = ⇒ ( h, k ) = ⎜ − , ⎟ ⎝ 14 7 ⎠ 13 −24 −14

Now the differential equation transforms to dY 6 X + 5Y = , which is a homogeneous equation. dX 4X +Y Putting Y = vX ; the equation becomes v+ X

Ch02.indd 26

dv 6 + 5v = dX 4+v

or

X

dv 6 + 5v 6 + 5v − 4 v − v 2 = −v= dX 4+v 4+v

12/9/2011 11:36:52 AM

Differential Equations of First Order and First Degree

2-27

dX v+4 + dv = 0 X v2 − v − 6

or

Putting into partial fractions 5

dX dv dv +7 −2 =0 X v −3 v+2

Integrating and dropping logs (Y − 3 X )7 = c1 (Y + 2 X )2 ⇒

1 1 (2 y − 6 x − 9)7 = c1 2 (7 y + 14 x + 1)2 7 2 7

Let

v+4 A B ≡ + v −v −6 v −3 v+2

A=

v+4 7 = v + 2 v =3 5

2

Y − 3X =

B=

2 y − 6x − 9 2

v+4 2 =− v − 3 v = −2 5

Y + 2X =

7 y + 14 x + 1 7

Absorbing the constants in the arbitrary constants C we can write the solution as (2 y − 6 x − 9)7 = C (7 y + 14 x + 1)2 Alternative method (Method of proportions) By the property of proportions and by using (1, λ ) as multipliers, we can write the differential equation as d ( y + λ x) dy dx = = 6 x + 5 y − 3 4 x + y + 2 (6 + 4 λ ) x + (5 + λ ) y − 3 + 2λ Choose λ so that λ =

6 + 4λ ⇒ λ 2 + λ − 6 = ( λ + 3)( λ − 2) = 0. 5+ λ

For λ = −3 and 2 we get from the last term d ( y − 3x ) d ( y + 2 x) = 2 y − 6 x − 9 7 y + 14 x + 1

Ch02.indd 27

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2-28

Differential Equations

Integrating, we have 1 1 log (2 y − 6 x − 9) = log (7 y + 14 x + 1) + const. 2 7 Dropping logs, after simplification, (2 y − 6 x − 9)7 = C (7 y + 14 x + 1)2 Example 2.1.30 Solve

„

dy 2 x + y − 2 = . dx 3 x + y − 3

Solution By the property of proportions, we have d ( y + λ x) dy dx = = 2 x + y − 2 3 x + y − 3 (2 + 3λ ) x + (1 + λ ) y − (2 + 3λ ) 2 + 3λ ⇒ λ 2 − 2λ − 2 = 0 ⇒ λ1 , λ 2 = 1 ± 3 Choose l such that λ = λ 1 + Each ratio =

d ( y + λ2 x) 1 d ( y + λ1 x ) 1 = ⋅ 1 + λ1 y + λ1 x − λ1 1 + λ 2 y + λ 2 x − λ 2

Integrating we get 1 1 ⋅ log [ y + λ1 ( x − 1)] = ⋅ log [ y + λ 2 ( x − 1)] + const. 1 + λ1 1 + λ2 ⇒ [ y + λ1 ( x − 1)]2 −

3

= c [ y + λ 2 ( x − 1)]2+

where λ1 = 1 + 3, λ 2 = 1 − 3 and Example 2.1.31 Solve

3

1 1 = 2 − 3, = 2 + 3. „ 1 + λ1 1 + λ2

dy 6 x + 2 y + 7 = . dx x + 2y − 3

Solution By the property of proportions and by using (1, l) as multipliers we have d ( y + λ x) dy dx = = + + + − + + 6 x 2 y 7 x 2 y 3 (6 λ ) x (2 + 2λ ) y + 7 − 3λ We choose l so that λ = (6 + λ ) (2 + 2λ ) ⇒ 2λ 2 + λ − 6 = 0 ⇒ (2λ − 3) ( λ + 2) = 0

Ch02.indd 28

⇒

λ1 = 3 2,

λ 2 = −2

12/9/2011 11:36:53 AM

Differential Equations of First Order and First Degree

2-29

3 Substituting the two values λ1 = and λ 2 = −2 for l we get 2 3 d ( y + x) d ( y − 2 x) 2 = 3 3 (6 + ) x + 5 y + (7 − 3. ) (6 − 2) x + (2 − 2.2) y + 7 − 3( −2) 2 2 d (2 y + 3 x ) 5 d ( y − 2 x) ⇒ + =0 3 x + 2 y + 1 2 ( y − 2 x ) − 13 Integrating and dropping logs we obtain (2 y + 3 x + 1)2 (2 y − 4 x − 13)5 = C Example 2.1.32 Solve

„

dy − x + 3 y + 1 = . dx x+ y+2

Solution Here the method of proportions is not applicable since we obtain roots λ1 = λ 2 = −1 for λ. We apply the usual method of case (B). Putting x = X + h, y = Y + k and determining h, k from

}

− h + 3k + 1 = 0 ⎛ 5 3⎞ we have ( h, k ) = ⎜ − , − ⎟ h+k +2= 0 ⎝ 4 4⎠ dY X + 3Y =− . dX X +Y dy dv −1 + 3v dv −1 + 2v − v 2 =v+x = or x = . Put y = vx ⇒ dx dx 1+ v dx 1+ v Separating the variables and integrating, after multiplying by 2, Then the given equation reduces to

2∫

dx 2 + 2v dx 2v − 2 4 + dv = 0 ⇒ 2∫ + ∫ 2 dv + ∫ dv = 0 x ∫ v 2 − 2v + 1 x v − 2v + 1 ( v − 1)2

we get 2x 4 y−x = constant ⇒ y − x = c1 e log x + log ( v − 1) − v −1 5 3⎞ 3⎞ ⎛ 5⎞ 1 ⎛ ⎛ But ( X ,Y ) = ⎜ x + , y + ⎟ ; Y − X = ⎜ y + ⎟ − ⎜ x + ⎟ = y − x − . ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 4 4 4 4 2 2

2

The required solution is ⎡ 4x + 5 ⎤ (2 y − 2 x − 1) = C exp ⎢ ⎥ ⎣ 2 y − 2 x − 1⎦

Ch02.indd 29

„

12/9/2011 11:36:53 AM

2-30

Differential Equations

EXERCISE 2.3 Solve the following differential equations: 1.

2.

dy 3 x − 6 y + 3 . = dx x − 2y +1 Ans: y = 3 x + C dy 3 x + 2 . = dx 2 y − 3 Ans: 3 x 2 − 2 y 2 + 4 x + 6 y + C = 0

3.

4.

5.

dy y+3 = . dx 2 x − 7 Ans: ( y + 3)2 = C (2 x − 7) dy 9 = . dx 5 x − 7 y + 13 Ans: (25 x − 35 y + 2)9 = C e 5 y dy 2 x − 5 y − 6 = . dx 11 Ans: (10 x − 25 y − 52)11 = C e −5 x

6.

7.

8.

9.

dy x − y + 2 . = dx x + y − 1 Ans: y 2 + 2 xy − x 2 − 2 y − 4 x + C = 0 dy 2 x − y + 1 = . dx x + 2 y − 3 Ans: y 2 + xy − x 2 − x − 3 y + C = 0 dy 2 x + y + 6 = . dx − x + y − 3 Ans: 2 x 2 + 2 xy − y 2 + 12 x + 6 y + C = 0 dy 2x + 3y + 1 = . dx −3 x + 4 y − 1 Ans: x 2 + 3 xy + 2 y 2 + x − y + C = 0

Ch02.indd 30

12/9/2011 11:36:54 AM

Differential Equations of First Order and First Degree

10.

11.

12.

13.

14.

dy y−x = . dx y − x + 2 Ans: x 2 + 2 xy − y 2 − 4 y + C = 0 dy x − 3 y + 4 . = dx 3 x − 9 y − 7 Ans: x 2 − 6 xy + 9 y 2 + 8 x + 14 y + C = 0 dy x + 2 y − 3 = . dx 2 x + y − 3 Ans: ( y − x )3 = C ( y + x − 2) dy ax + by − a = . dx bx + ay − b Ans: ( y − x + 1) a + b ( y + x − 1) a − b = C dy 2 x + y − 2 = . dx 3 x + y − 3 Ans: [ y + (1 + 3)( x − 1)](2 −

15.

2-31

3)

= C [ y + (1 − 3)( x − 1)](2+

3)

dy x + y + 1 = . dx x + y − 1 Ans: y = x + log( x + y ) + c

16. (2 x 2 + 3 y 2 − 7) x dx + (3 x 2 + 2 y 2 − 8) ydy = 0. [Hint: Put x 2 = X , y 2 = Y and group the terms: (2 X − 7)dX + 3d ( XY ) + (2Y − 8)dY = 0] Ans: X 2 + 3 XY + Y 2 − 7 X − 8Y + C = 0 or x 4 + 3 x 2 y 2 + y 2 − 7 x 2 − 8 y 2 + C = 0 17.

18.

dy (2e x − e y + 1)e x = . dx (e x + 2e y )e y dv 2u − v + 1⎤ ⎡ x y ⎢⎣Hint: Put e = u, e = v; du = u + 2v ⎥⎦ Ans: u 2 − uv − v 2 + u + C = 0 or e 2 x + e x = C + e x + y + e 2 y dy ax + by + g + = 0. dx hx + by + f Ans: 2 x 2 + 2hxy + by 2 + 2 gx + 2 fy + C = 0

Ch02.indd 31

12/9/2011 11:36:55 AM

2-32

Differential Equations

2.1.4 Exact Equations Definition 2.1.33 A differential equation which is obtained from its primitive by mere differentiation without any further operation is called an exact equation. xdy − ydx The equations xdy + ydx = 0, xdx + ydy = 0 and =0 x2 are exact since these can be written as d ( xy ) = 0, 12 d ( x 2 + y 2 ) = 0 and ⎛ y⎞ d ⎜ ⎟ = 0, respectively. ⎝ x⎠ The following theorem gives a criterion for exact equations. Theorem 2.1.34 If M ( x, y ) and N ( x, y ) are real-valued functions having continuous first partial derivatives on some rectangle R : x − x0 ≤ a, y − y0 ≤ b, then a necessary and sufficient condition for the equation (2.48) Mdx + Ndy = 0 to be exact in R is that ∂M ∂N = ∂y ∂x

in R.

(2.49)

Proof (i) The condition is necessary: Suppose Eq. (2.48) is exact. Then there exists a function u( x, y ) such that ∂u ∂u dx + dy ∂x ∂y = Mdx + Ndy

du =

(2.50)

∂u ∂u , N= (2.51) ∂x ∂y Differentiating M and N partially with respect to y and x, respectively ∂M ∂ ⎛ ∂u ⎞ ∂ 2 u (2.52) = ⎜ ⎟= ∂y ∂y ⎝ ∂x ⎠ ∂y ∂x ⇒M =

∂N ∂ ⎛ ∂u ⎞ ∂ 2 u = ⎜ ⎟= ∂x ∂x ⎝ ∂y ⎠ ∂y ∂x

(2.53)

From Eqs. (2.52) and (2.53) it follows that ∂M ∂N = ∂y ∂x

Ch02.indd 32

⎛ ∂2u ∂2u ⎞ ⎜⎝∵ ∂y∂x = ∂x∂y ⎟⎠

(2.54)

12/9/2011 11:36:55 AM

Differential Equations of First Order and First Degree

2-33

(ii) The condition is sufficient: Assume that Eq. (2.49) holds. Let

∫

v( x, y ) =

Mdx

y constant

where the integral is evaluated partially with respect to x treating y as ∂v constant so that = M. ∂x ∂N ∂M ∂ ⎛ ∂v ⎞ (2.55) Now, = = ⎜ ⎟ ∂x ∂y ∂y ⎝ ∂x ⎠ ∂2 v ∂ ⎛ ∂v ⎞ = ∂y∂x ∂x ⎜⎝ ∂y ⎟⎠ ∂ ⎛ ∂v ⎞ ⇒ N − ⎟ =0 ⎜ ∂x ⎝ ∂y ⎠ =

∂v = φ ( y ), a function of y alone ∂y ∂v ⇒N = + φ ( y) ∂y Hence, the equation becomes ⎛ ∂v ⎞ ∂v 0 = Mdx + Ndy = dx + ⎜ + φ ( y )⎟ dy ∂x ⎝ ∂y ⎠ ⇒N −

(

= d v + ∫ φ ( y )dy

)

(2.56)

which proves that the equation is exact. Procedure for solving an exact equation (1) Test if ∂M = ∂N . ∂y ∂x (2) Compute v =

∫

Mdx

y constant

u=

(or)

∫

Ndy.

x constant

(3) Compute (or)

Ch02.indd 33

⎛

∂v ⎞

∫ ⎜⎝ N − ∂y ⎟⎠ dy ⎛

∂u ⎞

∫ ⎜⎝ M − ∂x ⎟⎠ dx.

12/9/2011 11:36:56 AM

2-34

Differential Equations

(4) The general solution is ⎛ ∂v ⎞ Mdx + ∫ ⎜ N − ⎟ dy = c ∂y ⎠ ⎝ y constant

∫

∂u ⎞ ⎛ Ndy + ∫ ⎜ M − ⎟ dx = c. ⎝ ∂x ⎠ x constant

∫

(or)

„

Note 2.1.35 In most cases, collecting and grouping terms which are exact differentials will not only prove the exactness of the equation but also yield the general solution, on integration, as the illustrative examples will show. Note 2.1.36 The following formula

∫

y constant

Mdx + ∫ (terms of N

not containing x )dy = c (2.57) though in many cases, gives the correct solution, fails in some cases and is therefore not advisable. (Example 2.1.37 shows the failure of this method to give the correct solution.) Example 2.1.37 Solve y sin 2 x dx = ( y 2 + cos 2 x )dy. Solution Here,

M = y sin 2 x; N = − y 2 − cos 2 x ∂M ∂N = sin 2 x; = −2cos x( − sin x ) ∂y ∂x = sin 2 x ∂M ∂N = The equation is exact, since ∂y ∂x y Now v = ∫ Mdx = ∫ y sin 2 xdx = − cos 2 x 2 y constant y constant ⎛ 1 ∂v ⎞ 2 2 ⎜⎝ N − ∂y ⎟⎠ = − y − cos x + 2 cos 2 x 1 = − y 2 − cos 2 x + (2cos 2 x − 1) 2 1 = − y2 − 2

Ch02.indd 34

12/9/2011 11:36:56 AM

Differential Equations of First Order and First Degree

2-35

⎛ ∂v ⎞ 1 3 1 ⎜⎝ N − ∂y ⎟⎠ dy = − 3 y − 2 y Hence, the general solution is 1 1 y − cos 2 x − y 3 − y = constant 2 3 2 1 3 2 ⇒ y cos x + y = c 3 1 3 y ⇒ (cos 2 x + 1) + y = c. 2 3 Note 2.1.38 If we compute the solution using Eq. (2.57) we get y y y3 − cos 2 x − ∫ y 2 dy = constant ⇒ cos 2 x + =c 2 2 3 1 in which the term y is missing. 2 dy 3 x − 2 y + 5 Example 2.1.39 Solve + = 0. dx −2 x + 4 y + 1 Solution In the standard form, the equation is (3 x − 2 y + 5)dx + ( −2 x + 4 y + 1)dy = 0 M = 3 x − 2 y + 5; N = −2 x + 4 y + 1 My = −2 = Nx = −2

∫

„

The equation is exact since ∂M ∂y = ∂N ∂x v=

∫

y constant

=

Mdx = ∫ (3 x − 2 y + 5) dx

3 2 x − 2 xy + 5 x 2

∂v = −2 x ∂x ∂v N− = ( −2 x + 4 y + 1) − ( −2 x ) = 4 y + 1 ∂y ⎛ ∂v ⎞ ∫ ⎜⎝ N − ∂y ⎟⎠ dy = 2 y 2 + y The general solution is ⎛ 3 ∂v ⎞ v + ∫ ⎜ N − ⎟ dy = x 2 − 2 xy + 5 x + 2 y 2 + y 2 ∂y ⎠ ⎝ = Constant ⇒ 3 x 2 − 4 xy + 4 y 2 + 10 x + 2 y + c = 0.

Ch02.indd 35

12/9/2011 11:36:56 AM

2-36

Differential Equations

Alternative method: Grouping the terms (3 x + 5)dx − 2( ydx + xdy ) + (4 y + 1)dy = 0 ⎛3 ⎞ d ⎜ x 2 + 5 x ⎟ − 2d ( xy ) + d (2 y 2 + y ) = 0 ⎝2 ⎠ This shows that the equation is exact. Integrating it, we get

or

3 2 x + 5 x − 2 xy + 2 y 2 + y = Constant 2 3 x 2 − 4 xy + 4 y 2 + 10 x + 2 y + c = 0.

or

„

dy 3 x 2 − 2 xy − 5 . = dx x 2 + y 2 − 2 y Solution Writing the equation in the standard form (3 x 2 − 2 xy − 5)dx − ( x 2 + y 2 − 2 y )dy = 0 M = 3 x 2 − 2 xy − 5; N = − x 2 − y 2 + 2 y M y = −2 x = N x = −2 x

Example 2.1.40 Solve

The equation is exact, v= ∫

Mdx = x 3 − x 2 y − 5 x;

y constant

∂v = − x2 ∂y ⎛

∂v ⎞

∫ ⎜⎝ N − ∂y ⎟⎠ dy = ∫ ( − y

2

+ 2 y ) dy =

1 3 y + y2 3

The general solution is 1 3 y + y 2 + constant = 0 3 ⇒ 3 x 2 − y 3 − 3 x 2 y + 3 y 2 − 15 x + c = 0.

x3 − x2 y − 5x −

Alternative method: Grouping the terms (3 x 2 − 5)dx + (2 y − y 2 )dy − (2 xydx + x 2 dy ) = 0 1 ⎞ ⎛ ⇒ d ( x3 − 5x) + d ⎜ y3 − y3 ⎟ − d ( x 2 y) = 0 ⎝ 3 ⎠ The equation is exact; integrating and multiplying by 3, we get the general solution as 3 x 3 − 3 x 2 y − y 3 + 3 y 2 − 15 x + c = 0. „

Ch02.indd 36

12/9/2011 11:36:57 AM

Differential Equations of First Order and First Degree

2-37

x ⎛ x⎞ x Example 2.1.41 Solve 1 + e y dx + ⎜1 − ⎟ e y dy = 0. [JNTU 2000] y⎠ ⎝ Solution Here,

(

)

x ⎛ x⎞ x M = 1 + e y ; N = ⎜1 − ⎟ e y y⎠ ⎝ ⎡⎛ x ⎛ x⎞ x⎞ 1 1⎤ x x x M y = e y ⎜ − ⎟ = N x ⎢⎜1 − ⎟ − ⎥ e y = − 2 e y y⎠ y y⎦ y ⎝ y⎠ ⎣⎝ The equation is exact.

v=

∫

x ⎛ x⎞ x ⎛ ∂v x⎞ x = 1⋅ e y + y ⎜ − 2 ⎟ e y = ⎜1 − ⎟ e y ∂y y⎠ ⎝ y ⎠ ⎝

x

Mdx = x + ye y ;

y constant

∂v ⎛ x⎞ x ⎛ x⎞ x N− = 1 − ⎟ e y − ⎜1 − ⎟ e y = 0 ∂y ⎜⎝ y⎠ y⎠ ⎝ x

The general solution is x + ye y = c. Alternative method: Grouping the terms x ⎛ x ⎞ x dx + e y dy + ⎜ dx − dy⎟ e y = 0 y ⎠ ⎝ x x ⎛ x⎞ ⇒ dx + e y dy + y ⋅ e y d ⎜ ⎟ = 0 ⎝ y⎠

( ) x

⇒ dx + d ye y = 0 The equation is exact. x The general solution is x + ye y = c.

„

Example 2.1.42 Solve ( y 2 − 2 xy )dx = ( x 2 − 2 xy )dy. [JNTU 1995] Solution M = y 2 − 2 xy; N = 2 xy − x 2 M y = 2 y − 2x = N x = 2 y − 2x The equation is exact. ∂v v = ∫ Mdx = xy 2 − x 2 y; = 2 xy − x 2 ∂y y constant ∂v N− = − x 2 + 2 xy − 2 xy + x 2 = 0 ∂y The general solution is xy 2 − x 2 y = c.

Ch02.indd 37

12/9/2011 11:36:57 AM

2-38

Differential Equations

Alternative method: Grouping the terms y 2 dx + 2 xydy = x 2 dy + 2 xydx ⇒

d ( xy 2 ) − d ( x 2 y ) = 0

(The equation is exact.) The general solution is xy 2 − x 2 y = c. Example 2.1.43 Solve xdx + ydy = Solution LHS =

„

xdy − ydx . x2 + y2

1 d( x2 + y2 ) 2

( RHS =

xdy − ydx x2

(

x2 + y2 x

)

)

(dividing the numerator and denominator by x 2 ) =

d ( xy ) y⎞ ⎛ = d ⎜ tan −1 ⎟ . y 2 ⎝ x⎠ 1+ ( x )

From the above results, we observe that the equation is exact. Its general solutions is y x 2 + y 2 = 2 tan −1 + c. x „ Example 2.1.44 Solve (e y + 1) cos x dx + e y sin x dy = 0. Solution (e y cos xdx + e y sin xdy ) + cos xdx = 0 ⇒

d (e y sin x ) + d (sin x ) = 0

(The equation is exact.) Integrating, we get the general solution as (e y + 1)sin x = c.

Ch02.indd 38

„

12/9/2011 11:36:58 AM

Differential Equations of First Order and First Degree

2-39

EXERCISE 2.4 Solve: 1. (2 x + 3 y + 4)dx + (3 x − 6 y + 2)dy = 0. Ans: x 2 + 3 xy − 3 y 2 + 4 x + 2 y + c = 0 2. (3 x 2 − 9 x 2 y 2 + 2 xy ) dx + (6 y 2 − 6 x 3 y 2 + x 2 )dy = 0. Ans: x 3 + 2 y 3 − 3 x 3 y 2 + x 2 y = c 3. e x (sin x + cos x )sec y dx + e x sin x sec y tan y dy = 0. Ans: e x sin x sec y = c 4. (cos x cos y − cot x ) dx = sin x sin y dy. Ans: sin x cos y = log c sin x 1 5. { y (1 + x ) + sin y} dx + {x + log x + x cos y} dy = 0. Ans: xy + y log x + x sin y = c

6. 4 y sin 2 xdx + (2 y + 3 − 4 cos 2 x )dy = 0. Ans: y 2 + y − 2 y cos 2 x = c xdy − ydx = 0. 7. xdx + ydy + 2 x − y2 x+ y

Ans: ( x 2 + y 2 )e x − y = c 8. ( x 2 − ay )dx = ( ax − y 2 )dy. Ans: x 3 + y 3 = 3axy + c 9.

xdx + ydy xdy − ydx + 2 = 0. x2 + y2 x − y2 Ans: ( x 2 + y 2 )

10.

( )=c x+ y x− y

dy y cos x + sin y + y + = 0. dx sin x + x cos y + x Ans: xy + y sin x + x sin y = c

11.

y( xy + e x )dx − e x dy = 0. y2 Ans:

Ch02.indd 39

x2 2

⎡ ⎤ ye x dx − e x dy xdx + = 0.⎥ Hint : ⎢ 2 y ⎣ ⎦

+ ey = c x

12/9/2011 11:36:58 AM

2-40

12.

Differential Equations

ydx − xdy + xe x dx = 0. y2 Ans: xy + ( x − 1)e x = c

2.1.5 Inexact Equation—Reducible to Exact Equation by Integrating Factors Integrating factor (I.F.) If the differential equation Mdx + Ndy = 0 becomes exact when we multiply it by a function μ ( x, y ) then μ ( x, y ) is called an integrating factor of the equation. Consider the equation ydx − xdy = 0 Here ∂M ∂N M = y, N = − x ⇒ =1≠ = −1 ∂y ∂x 1 The equation is not exact. If we multiply it by 2 we get y 1 x dx − 2 dy = 0. y y Now M = 1y , N = − yx2 so that ∂M 1 ∂N =− 2 = . ∂y y ∂x So, the equation becomes exact. ily check that

1 x2

1 xy

, ,

1 x2 + y2

1 y2

is an integrating factor. We can eas-

are also integrating factors for the equation.

Integrating factor can be found by inspection, after grouping of terms. Table 2.1 gives the list of integrating factors. Example 2.1.45 Solve the following differential equations after finding the integrating factor in each case: Differential Equation (i) ydx − xdy + y 2 xe x dx = 0 (ii) ydx − xdy + 2 x 2 y sin x 2 dx = 0

Ch02.indd 40

Integrating Factor 1 y2 1 xy

12/9/2011 11:36:59 AM

Ch02.indd 41

2 xdx + 2 ydy = d [log( x 2 + y 2 )] x2 + y2

d( x2 + y2 ) xdy − ydx ⎛ y⎞ =d⎜ ⎟ ⎝ x⎠ x2

2 x2 + y2

2 1 x2

xdx + ydy

xdx + ydy

xdx + ydy

xdy − ydx

xdy − ydx

4.

5.

6.

7.

8.

−

1 y2

⎛ x⎞ ydx − xdy =d⎜ ⎟ ⎝ y⎠ y2 (Continued)

2 xdx + 2 ydy ⎡ ( x 2 + y 2 )1−a ⎤ =d⎢ ⎥ ( a ≠ 1) 2 2 a (x + y ) 1− a ⎣ ⎦

xdy + ydx ⎛ ( xy )1−a ⎞ =d ⎜ ( a ≠ 1) ⎝ 1 − a ⎟⎠ ( xy ) a

1 ( xy ) a

xdy + ydx

3.

2 + y 2 )a

xdy + ydx = d (log xy ) xy

1 xy

xdy + ydx

2.

( x2

d ( xy )

1

xdy + ydx

1.

Exact differential

Integrating factor

Group of terms

S. No

Table 2.1

Differential Equations of First Order and First Degree 2-41

12/9/2011 11:36:59 AM

Ch02.indd 42

1 x2 + y2

1 − 2 x + y2 y x2 −

2 x2 − y2

xdy − ydx

xdy − ydx

2xdy − ydx

xdy − 2 ydx

xdy − ydx

10.

11.

12.

13.

14.

x y2

1 xy

xdy − ydx

9.

2

( x y

)

)

xdy − ydx ⎛ 2 xdy − 2 ydx 2 x2 x + y⎞ = = d ⎜ log y 1 − ( x )2 x2 − y2 x − y ⎠⎟ ⎝

(

⎛ x2 ⎞ 2xydx − x 2 dy = d⎜ ⎟ 2 y ⎝ y⎠

⎛ y2 ⎞ ⎜⎝ ⎟⎠ x

= d tan −1

2xydy − y 2 dx =d x2

1 + xy

( )

ydx − xdy y2

y xdy − ydx x2 −1 = y 2 = d (tan x ) x2 + y2 1+ ( ) x

xdy − ydx

dy dx y⎞ ⎛ − = d ⎜ log ⎟ ⎝ y x x⎠

Exact differential

Table 2.1 (Continued) Integrating factor

Group of terms

S. No

2-42 Differential Equations

12/9/2011 11:37:00 AM

Differential Equations of First Order and First Degree

1 x2

(iii) ydx − xdy + 2 x 2 ye y2 dy = 0 (iv) ydx − xdy + ( x 2 + y 2 )

dx 1 − x2

2-43

=0

1 x + y2 2

(v) xdx + ydy + ( x 2 + y 2 ) tan xdx = 0

1 x + y2

(vi) y(1 + xy )dx + x(1 − xy )dy = 0

1 ( xy )2

2

Solution The suitable integrating factor in each case is shown against each differential equation. After multiplying by the integrating factor, we can derive the general solution as follows. (i) ydx − xdy + xe x dx = 0 y2

(ii)

⎛ x⎞

⇒

∫ d ⎜⎝ y ⎟⎠ + ∫ xe dx = c

⇒

x + ( x − 1) e x = c. y

x

ydx − xdy + 2 x sin x 2 dx = 0 xy dx dy − + sin( x 2 ) d ( x 2 ) = 0 x y x ⇒ log = cos x 2 + c. y ⇒

(iii)

ydx − xdy 2 + e y dy 2 = 0 x2 ⇒ ⇒ ⇒

Ch02.indd 43

⎛ y⎞ 2 − d ⎜ ⎟ + d (e y ) = 0 ⎝ x⎠ y − + ey = c x y ey = c + . x

12/9/2011 11:37:00 AM

2-44

(iv)

Differential Equations

ydx − xdy dx + =0 x2 + y2 1 − x2 ⇒ ⇒

ydx − xdy x2 y 2

+

() −d ( ) + 1+ ( ) 1+

x

y x

y 2 x

dx =0 1 − x2 dx =0 1 − x2

⇒ sin −1 x = tan −1 (v)

y + c. x

2 xdx + 2 ydy + tan xdx = 0 x2 + y2 ⇒ ⇒

d (log( x 2 + y 2 )) + d (log(sec x )) = 0 ( x 2 + y 2 )(sec x ) = c.

(vi) ydx + xdy + xy( ydx − xdy ) = 0 d ( xy ) dx dy + − =0 ( xy )2 x y 1 x ⇒ − + log = c. xy y ⇒

„

EXERCISE 2.5 Find the integrating factors by inspection and solve the following: 1. xdy − ( y − 3 x 2 ) dx = 0. Ans: 2. ye − x Ans:

y + 3x = c x + e − x dy = 2 xy 2 dx. e− x y

= x2 + c

3. xdy + 2 ydx = 2 y 2 xdy. Ans: xy 2 = ce y

2

4. ( x 2 + y 2 )( xdy − ydx ) = ( x 2 − y 2 ) ( xdx + ydy ). Ans: ( x + y ) = c( x − y ) ( x 2 + y 2 )

Ch02.indd 44

12/9/2011 11:37:01 AM

Differential Equations of First Order and First Degree

2-45

5. 2 xydy − y 2 dx + x 3e x dx = 0. Ans: y 2 + x ( x − 1) e x + cx = 0 Given below are the rules for finding integrating factors of Mdx + Ndy = 0 depending on the nature of functions M and N.

Rule 1 If Mdx + Ndy = 0 is a homogeneous differential equation and Mx + Ny ≠ 0, then Mx1+ Ny is an integrating factor of the equation. Note 2.1.46 Recall that we have already solved homogeneous differential equations reducing them to variables separable form by putting y = vx or x = vy. Rule 1 Provides an alternative method for its solution. Note 2.1.47 If Mx + Ny = 0, then

M y

=

ydx − xdy = 0 whose general solution is

N −x x y

and the equation becomes = c.

Rule 2 If Mdx + Ndy = 0 is of the form f1 ( xy ) ydx + f 2 ( xy ) xdy = 0 and Mx − Ny ≠ 0, then equation.

(

is an integrating factor of the

1 Mx − Ny

)

∂M ∂N Rule 3 If ∂y − ∂x / N = p( x ), purely a function of x alone, then e ∫ p ( x ) dx is an integrating factor of the equation Mdx + Ndy = 0.

Note that the linear equation dy where P, Q are funcdx + Py = Q tions of x alone may be written in the form Mdx + Ndy = 0 where M = Py − Q and N = 1 so that ∂M ∂y

− ∂∂Nx N

P −0 1 = P, =

p ( x ) dx , purely a function of x and the integrating factor in this case is e ∫ as we will see in linear equations discussed below.

Rule 4 If e∫

Ch02.indd 45

q ( y ) dy

(

∂N ∂x

)

− ∂∂My / M = q( y ), purely a function of y alone then

is an integrating factor of the equation Mdx + Ndy = 0.

12/9/2011 11:37:01 AM

2-46

Differential Equations

Rule 5 If Mdx + Ndy = 0 is expressible in the form x a y b ( mydx + nxdy ) + x c y d ( pydx + qxdy ) = 0 where a,b,c,d,m,n,p,q are all constants such that mq − np ≠ 0, then x h y k is an integrating factor of the equation for some suitable constants h, k to be determined from the two equations a +mh+1 = b +nk +1 and c + h +1 = d +qk +1 p Example 2.1.48 Solve ( x 2 + y 2 )dx − xydy = 0 by finding an integrating factor. Solution This is a homogeneous differential equation. Here, M = x2 + y2 , N = − xy; ∂M ∂N = 2 y, = −y ∂y ∂x Since

∂M ∂N ≠ , the equation is not exact ∂y ∂x Mx + Ny = ( x 2 + y 2 ) x − ( xy ) y = x 3 ≠ 0

Multiplying by the integrating factor, 1 1 = Mx + Ny x 3 y ⎛ 1 y2 ⎞ ⎜⎝ + 3 ⎟⎠ dx − 2 dy = 0 x x x ⎛ 1 y3 ⎞ v = ∫ Mdx = ∫ ⎜ + 3 ⎟ dx ⎝x x ⎠ y constant 2 y = log x − 2 2x ∂v y ∂v =− 2; N− =0 ∂y x ∂y ∴

The general solution is log x =

y2 + c. 2x2 „

Ch02.indd 46

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Differential Equations of First Order and First Degree

2-47

Example 2.1.49 Find an integrating factor and solve dy y x 2 + y 2 = + . dx x x2 Solution Putting the equation in the standard form ( xy + x 2 + y 2 ) dx − x 2 dy = 0 M = xy + x 2 + y 2 , ∂M = x + 2 y, ∂y Since

N = − x2 ∂N = −2 x ∂x

∂M ∂N ≠ , the equation is not exact ∂y ∂x Mx + Ny = x ( xy + x 2 + y 2 ) − x 2 y = x ( x2 + y2 ) ≠ 0

The given equation is a homogeneous differential equation. ∴

Integrating factor =

1 1 = 2 Mx + Ny x( x + y 2 )

Multiplying the equation by the integrating factor dx xdy − ydx = 2 x x + y2 xdy − ydx

⇒ ⇒

d ( xy ) dx x2 = = x 1 + ( xy )2 1 + ( xy )2 y log x = tan −1 + c. x

„

Example 2.1.50 Solve ( xy + 1) ydx + ( − xy + 1) xdy = 0. Solution This equation is of the form f1 ( xy ) ydx + f 2 ( xy ) xdy = 0

Ch02.indd 47

12/9/2011 11:37:03 AM

2-48

Differential Equations

By Rule 2, the integrating factor is

1 Mx − Ny

if Mx − Ny ≠ 0.

Mx − Ny = ( xy + 1) yx − ( − xy + 1) yx = 2 x 2 y 2 ≠ 0 Multiplying by the integrating factor = ten as

1 x2 y2

, the equation can be writ-

dx dy d ( xy ) 1 − + = 0 ⇒ log x /y = + c. 2 x y ( xy ) xy

„

Example 2.1.51 Solve ( x 2 y 2 + xy + 1) ydx + ( x 2 y 2 − xy + 1) xdy = 0. Solution The equation is of the form f1 ( xy ) ydx + f 2 ( xy ) xdy = 0 Mx − Ny = x 3 y 3 + x 2 y 2 + xy − ( x 3 y 3 − x 2 y 2 + xy ) = 2x2 y2 ≠ 0 Integrating factor =

Multiplying the equation by

1 1 = 2 2 Mx − Ny 2 x y

1 and omitting the constant, x y2 2

⎛ ⎛ 1 1 ⎞ 1 1 ⎞ ⎜⎝1 + xy + x 2 y 2 ⎟⎠ ydx + ⎜⎝1 − xy + x 2 y 2 ⎟⎠ xdy = 0 ⎛ 1 ⎞ 1 ⎜⎝1 + x 2 y 2 ⎟⎠ ( ydx + xdy ) + xy ( ydx − xdy ) = 0 ⎛ 1 ⎞ 1 1 ⎜⎝1 + x 2 y 2 ⎟⎠ d ( xy ) + x dx − y dy = 0 ⎡ ⎛ x⎞⎤ ⎛ 1 ⎞ ⎜⎝1 + ( xy )2 ⎟⎠ d ( xy ) + d ⎢log ⎜⎝ y ⎟⎠ ⎥ = 0 ⎣ ⎦ The general solution is xy −

Ch02.indd 48

1 x + log = c. xy y

„

12/9/2011 11:37:03 AM

Differential Equations of First Order and First Degree

2-49

Example 2.1.52 Solve ( x − y ) dx − dy = 0. Solution M = x − y, N = −1, ∂M ∂N = −1, = 0; ∂y ∂x ∂M ∂N −1 − 0 ∂y − ∂x = = 1 = p ( x) N −1 (Constant can be considered as a function of x.) By Rule 3, the integrating factor is e ∫ p ( x ) dx = e ∫ 1dx = e x . Multiplying by the integrating factor = e x , the equation can be written as xe x dx = ( ye x dx + e x dy ) ⇒ ( x − 1) e x = ye x + c or x = y + 1 + ce − x .

„

Example 2.1.53 Solve (3 xy − 2 y 2 ) dx + ( x 2 − 2 xy ) dy = 0. Solution M = 3 xy − 2 y 2 , ∂M = 3x − 4 y, ∂y

∂M ∂y

− ∂∂Nx N

N = x 2 − 2 xy ∂N = 2x − 2 y ∂x ∂M ∂N ≠ ∂y ∂x

(3 x − 4 y ) − (2 x − 2 y ) x( x − 2 y ) x − 2y 1 = = = p ( x ), a function of x x( x − 2 y ) x =

By Rule 3, the integrating factor is e ∫ p ( x ) dx = e ∫ x dx = e log x = x. 1

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2-50

Differential Equations

Multiplying by integrating factor = x, the equation becomes (3 x 2 y − 2 xy 2 ) dx + ( x 3 − 2 x 2 y ) dy = 0 d ( x3 y) = d ( x 2 y 2 ) ⇒ x3 y = x 2 y 2 + c „

is the general solution of the differential equation. Example 2.1.54 Solve ( xy 3 + y ) dx + 2( x 2 y 2 + x + y ) dy = 0. Solution M = xy 3 + y ∂M = 3 xy 2 + 1 ∂y 1 ⎛ ∂N ∂M ⎞ − M ⎜⎝ ∂x ∂y ⎟⎠

N = 2x2 y2 + 2x + 2 y ∂N = 4 xy 2 + 2 ∂x xy 2 + 1 1 = = = q ( y ), 2 ( xy + 1) y y

a function of y alone. 1 By Rule 4, the integrating factor is e ∫ q ( y ) dy = e ∫ y dy = y. Multiplying by integrating factor, the equation can be written as ( xy 4 dx + 2 x 2 y 3 dy ) + ( y 2 dx + 2 xydy ) + 2 ydy = 0 1 d ( x 2 y 4 ) + d ( xy 2 ) + d ( y 2 ) = 0 2 General solution is 1 2 4 x y + xy 2 + y 2 = c. 2

„

Example 2.1.55 Solve ( y + xy 2 ) dx − xdy = 0. Solution M = y + xy 2 ; N = −x ∂M ∂N = 1 + 2 xy; = −1 ∂y ∂x 1 ⎛ ∂N ∂M ⎞ 1 ( −1 − 1 − 2 xy ) − = M ⎜⎝ ∂x ∂y ⎟⎠ y (1 + xy ) −2 (1 + xy ) 2 = = − = q ( y) y (1 + xy ) y ( xy ≠ −1)

Ch02.indd 50

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Differential Equations of First Order and First Degree

2-51

By Rule 4, the integrating factor is e ∫ q ( y ) dy = e − ∫ y dy = e −2log y = 2

1 , y2

a pure function of y. Multiplying the equation by d ( x2 ) + 2

(

ydx − xdy y2

1 y2

we can write the equation as

) = 0, whose general solution is 2x = c. y

x2 +

„

Example 2.1.56 Solve xy 3 ( ydx + 2 xdy ) + (3 ydx + 5 xdy ) = 0. Solution The given differential equation is xy 3 ( ydx + 2 xdy ) + (3 ydx + 5 xdy ) = 0 Here

M = xy 4 + 3 y

(2.58)

N = 2 x2 y3 + 5x

M y = 4 xy 3 + 3 N x = 4 xy 3 + 5 Since M y ≠ N x , the equation is not exact. We can easily verify that Rules 1–4 are not applicable here. So, we try to find an I.F. of the form x h y k, by applying Rule 5. Comparing Eq. (2.58) with the standard form x a y b ( mydx + nxdy ) + x c y d ( pydx + qydy ) = 0

(2.59)

We have

Also,

Ch02.indd 51

a = 1,

b = 3,

m = 1, n = 2

c = 0,

d = 0,

p = 3, q = 5

mp − nq = 1.3 − 2.5 = −7 ≠ 0

(2.60) (2.61)

12/9/2011 11:37:04 AM

2-52

Differential Equations

The constants h, k are determined from a + h +1 b + k +1 c + h +1 d + k +1 = and = m n p q 1+ h +1 3 + k +1 0 + h +1 0 + k +1 = and = 1 2 3 5 5 h + 5 = 3 k + 3 ⇒ 2h + 4 = k + 4 and ⇒ 2h = k and 5h + 2 = 3k (2.62) ⇒ ( h, k ) = (2, 4) (2.63) ⇒

An integrating factor is x 2 y 4 . Multiplying Eq. (2.58) by this integrating factor, we have x 3 − y 8 dx + 2 x 4 y 7 dy +3 x 2 y 5 dx + 5 x 3 y 4 dy = 0 (2.64) which, on regrouping and expressing as exact differentials, yields

⇒ ⇒

1 d ( x 4 y8 ) + d ( x3 y5 ) = 0 4 x 4 y8 + 4 x3 y5 = c x 3 y 5 ( xy 3 + 4) = c

which is the required solution.

(2.65) „

Example 2.1.57 Solve x(3 ydx + 2 xdy ) + 8 y 4 ( ydx + xdy ) = 0. Solution The given differential equation is x(3 ydx + 2 xdy ) + 8 y 4 ( ydx + xdy ) = 0

(2.66)

We observe that an integrating for this equation is of the form x h y k for some h, k . Multiplying Eq. (2.66) by x h y k, it can be written in the form

Mdx + Ndy = 0 where

(2.67)

M = 3 x h+1 y k +1 + 8 x h y k + 5, N = 2 x h+ 2 y k + 8 x h+1 y k + 4

∴

M y = 3( k + 1) x h+1 y k +8( k + 5) x h y k + 4

Ch02.indd 52

(2.68)

12/9/2011 11:37:04 AM

Differential Equations of First Order and First Degree

2-53

N x = 2( h + 2) x h+1 y k + 8( h + 1) x h y k + 4 Exactness condition is M y = N x ⇒ 3k = 2h + 1, k + 4 = h ⇒ ( h + k ) = (13,9) Now the left-hand side expression of Eq. (2.66) is exact

(2.69)

3 8 d ( x15 y10 ) + d ( x14 y14 ) = 0 15 14 ⇒ x14 y10 (7 x + 20 y 4 ) = c ⇒ Mdx + Ndy =

„

which is the required solution. Example 2.1.58 Solve 2 x 2 ( ydx + xdy ) + y( ydx − xdy ) = 0. Solution The given differential equation is 2 x 2 ( ydx + xdy ) + y ( dx − xdy ) = 0 Here ∴

(2.70)

M = 2 x 2 y + y 2 N = 2 x 3 − xy M y = 2x2 + 2 y N x = 6x2 − y

Since M y ≠ N x , the equation is not exact. To find an integrating factor of the form x h y k , we multiply Eq. (2.70) by x h y k so that new value of M = 2 x h+ 2 y k +1 + x hy and N = 2 x h+ 3 y k − x h +1 y k +1 M y = 2( k + 1) x h+ 2 y k + ( k + 2) x h y k +1 N x = 2( h + 3) x h + 2 y k − ( h + 1) x h y k +1 k +2

Equating M y = N x , we have k = h+2

k + 3 = −h

( h, k ) = ( −5 2, −1 2 ) ∴

Mdx + Ndy = (2 x −1 2 y1 2 + x −5 2 y 3 2 )dx + (2 x1 2 y −1 2 − x −3 2 y1 2 )dy =0 2 ⇒ 4 d ( x1 2 y1 2 ) − d ( x − 3 2 y 3 2 ) = 0 3

Ch02.indd 53

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2-54

Differential Equations

Integrating, we get 2 −3 2 3 2 x y = Constant 3 ⇒ 6 x1 2 y1 2 − x −3 2 y 3 2 = c 4 x1 2 y1 2 −

which is the required solution.

„

Example 2.1.59 Solve xy( ydx + xdy ) + x 2 y 2 (2 ydx − xdy ) = 0. Solution Multiplying the equation by an integrating factor x h y k, we obtain Mdx + Ndy = ( x h+1 y k + 2 + 2 x h + 2 y k + 3 ) dx + ( x h+ 2 y k +1 − x h + 3 y k + 2 ) dy (2.71) =0 My = Nx ⇒ ( k + 2) x h+1 y k +1 + 2( k + 3) x h+ 2 y ( k + 2) = ( h + 2) x ( h+1) y ( k +1) − ( h + 3) x h+ 2 y k + 2 Equating the coefficients of like powers k + 2 = h + 2, 2k + 6 = − h − 3 ⇒ ( h, k ) = ( −3, −3)

(2.72)

Now Eq. (2.71) becomes ( x −2 y −1 + 2 x −1 )dx + ( x −1 y −2 − y −1 )dy = 0 − d ( x −1 y −1 ) + d (2log x ) − log y = 0 Integrating, we get −

1 log x 2 + =c xy y

the required general solution.

„

EXERCISE 2.6 Solve the following equations by finding integrating factors: 1. ( x 2 y − 2 xy 2 )dx − ( x 3 − 3 x 2 y )dy = 0. (Madras 1975, Karnataka 1971, Calcutta Hon. 1975) Ans: ( x y ) + log ( y 3 x 2 ) = c

Ch02.indd 54

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Differential Equations of First Order and First Degree

2-55

2. ( x 2 + y 2 )dx − 2 xydy = 0. Ans: x 2 − y 2 = cx 3. x 2 ydx − ( x 3 + y 3 )dy = 0. 3 3 Ans: y = ce x / (3 y )

4. y( xy + 2 x 2 y 2 )dx + x( xy − x 2 y 2 )dy = 0. (Rajasthan 1969, Kanpur 1974, Karnataka 1971, Punjab 1971) 1 Ans: log( x 2 y ) = +c xy 5. ( x 3 y 3 + x 2 y 2 + xy + 1) ydx + ( x 3 y 3 − x 2 y 2 − xy + 1) xdy = 0. (Kanpur 1980, Gorakhpur 1972) 1 + log y 2 Ans: xy = c + xy 6. ( xy sin xy + cos xy ) ydx + ( xy sin xy − cos xy ) xdy = 0. (Gorakhpur 1974, Kanpur 1977, Lucknow 1975, Rajasthan 1976, Marathawada 1974) Ans: ( x /y )sec xy = c 7. ( x 2 + y 2 + 2 x )dx + 2 ydy = 0. (Srivenkateswara 1984, Calicut 1983) Ans: e x ( x 3 + y 3 ) = c 8. ( y + log x )dx − xdy = 0. (Marathawada 1994) Ans: ex + y + log x + 1 = 0 9. (3 x 2 y 4 + 2 xy )dx + (2 x 3 y 3 − x 2 )dy = 0. (Calcutta Hon. 1952, 1954; Utkal 1980) 1 Ans: x 3 y 3 + 2 x 3 10. ( xy + y )dx + 2( x 2 y 2 + x + y 4 )dy = 0. Ans: 3 x 2 y 4 + 6 xy 2 + 2 y 6 = c 11. x(4 ydx + 2 xdy ) + y 3 (3 ydx + 5 xdy ) = 0. Ans: x 4 y 2 + x 3 y 3 = c

Ch02.indd 55

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2-56

Differential Equations

12. xy 3 ( ydx + 2 xdy ) + (3 ydx + 5 xdy ) = 0. Ans:

1 4

x 4 y8 + x3 y5 = c

2.1.6 Linear Equations A differential equation of the form dy p0 ( x ) + p1 ( x ) y = p2 ( x ) ( p0 ( x ) ≠ 0) (2.73) dx where p0 , p1 and p2 are continuous functions of x in some interval I, is called a first order linear differential equation in y. Dividing Eq. (2.73) by p0 , we can write it as dy + P ( x) y = Q ( x) (2.74) dx which is taken as the standard form of the first order linear equaoccur tion. Here the dependent variable y and its derivative dy dx separately and to a first degree. For example, (i) (1 + x )

dy + 2 xy = 3 x 2 dx

dy + (sin x ) y = tan x dx If Q ( x ) ≡/ 0, then Eq. (2.73) is called a non-homogeneous linear equation. If Q ( x ) ≡ 0, then it is called a homogeneous or reduced linear equation. Writing Eq. (2.73) in the form (ii) cos x

( Py − Q ) dx + dy = 0

(2.75)

Mdx + Ndy = 0

(2.76)

and comparing it with we have M = Py − Q , N = 1. Since

(

∂M ∂y

− ∂∂Nx N

Ch02.indd 56

) = P ( x)

(2.77)

12/9/2011 11:37:06 AM

Differential Equations of First Order and First Degree

2-57

a function of x alone e ∫ pdx is an integrating factor of Eq. (2.74). Multiplying Eq. (2.74) by the integrating factor e ∫ pdx , we have dy ∫ pdx d + ye ∫ pdx = ( ye ∫ pdx ) = Qe ∫ pdx e dx dx Separating the variables and integrating, we get the general solution of Eq. (2.74) as ye ∫ pdx = c + ∫ Qe ∫ pdx dx (2.78) where c is an arbitrary constant. Method of solving a linear equation (1) Write the equation in the standard form dy + Py = Q. dx (2) Find the integrating factor e ∫ pdx and multiply the equation by the integrating factor. (3) Write the general solution of the differential equation as y ( integrating factor ) = c + ∫ Q (integrating factor ) dx, after evaluating the integral. Note 2.1.60 The words homogeneous and non-homogeneous used here are not to be confused with similar words used earlier. Note 2.1.61 Sometimes a linear equation may be written in the form dy Po ( x ) + P ( x ) y = Q ( x ) (2.79) dx where Po′ = P (2.80) so that Eq. (2.79) can be written as d ( Po y ) = Qdx which, on integration, yields the general solution Po y = c + ∫ Qdx. (2.81) dy + 2 xy = cot x. dx Solution This can be written as Example 2.1.62 Solve (1 + x 2 )

d ⎡(1 + x 2 ) y ⎤⎦ = cot x dx ⎣

Ch02.indd 57

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2-58

Differential Equations

Separating the variables and integrating

(1 + x 2 ) y = logsin x + c where c is an arbitrary constant.

„

Note 2.1.63 An equation which is not linear in y may be linear in the variable x. Example 2.1.64 Solve y 2 dx + ( xy − 1) dy = 0. Solution Rewriting this equation as dx 1 1 + x= 2, dy y y which is linear in x. Integrating factor = e ∫ y dy = e log y = y 1 x ( integrating factor ) = c + ∫ 2 y Integrating factor dy 1 ⇒ xy = c + ∫ dy = c + log y. y 1

„

dy 1 + y = e x + sin x. dx x Solution The equation is linear in y Example 2.1.65 Solve

Here

P ( x) =

1 , x

Q ( x ) = e x + sin x

Integrating factor = e ∫ pdx 1 = e ∫ x dx = e log x = x

Multiplying the equation by x, we have dy + y = x (e x + sin x ) dx d ⇒ ( xy ) = x (e x + sin x ) dx x

Ch02.indd 58

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Differential Equations of First Order and First Degree

2-59

xy = c + x (e x − cos x )

Integrating

= c + x (e x − cos x ) − e x + cos x xy = c + ( x − 1)(e x − cos x ).

„

dy + y = 2log x. dx Solution Writing the equation in the standard form

Example 2.1.66 Solve x log x

dy 1 2 + y= dx x log x x P ( x) =

Here

1 2 , Q ( x) = x log x x

This is a linear equation in y 1

1

∫ Pdx = ∫ x log x dx = ∫ log x d (log x ) = log (log x ) Integrating factor = e ∫ Pdx = e log (log x ) = log x. The general solution is 2 2 y ( log x ) = c + ∫ log xdx = c + 2∫ log xd ( log x ) = c + (log x ) x

[Hint :

put log x = t , 2 1 ∫ x log xdx x dx = dt = ∫ 2tdt = t 2 ⎤⎦

Example 2.1.67 Solve ( x + 2 y 3 )

„

dy = y. dx

Solution This equation can be written as y

Here

Ch02.indd 59

dx − x = 2 y 3 or dy

1 P ( y) = − , y

dx 1 − ⋅ x = 2 y2 dy y

Q ( y) = 2 y2

12/9/2011 11:37:08 AM

2-60

Differential Equations

The equation is linear in x 1

∫ P ( y ) dy = ∫ − y dy = − log y = log

1 y

Integrating factor = e ∫ P ( y )dy = e log y = 1

1 y

The general solution is x⋅

or

1 1 = c + ∫ 2 y 2 ⋅ dy y y = c + y2 x = c + y2 y

Example 2.1.68 Solve (1 + y 2 )

„

dx −1 + x = e tan y. dy

[JNTU 2001S].

Solution Writing the equation as dx 1 1 −1 + x= e tan y 2 1 + y2 dy 1 + y Here,

P ( y) =

1 1 −1 e tan y , Q ( y) = 2 2 1+ y 1+ y

The equation is linear in x 1

∫ P ( y ) dy = ∫ 1 + y

2

dy = tan −1 y

Integrating factor = e ∫ P ( y )dy = e tan

−1

y

The general solution is xe tan

Ch02.indd 60

−1

y

=c+∫

1 −1 e 2 tan dy 2 1+ y

12/9/2011 11:37:08 AM

Differential Equations of First Order and First Degree

1 dy = dt 1 + y2

Put tan −1 y = t ,

xe tan

−1

y

2-61

= c + ∫ e 2t dt 1 = c + e 2t 2 1 2(tan −1 y ) =c+ e . 2

„

dy + y = tan x. dx Solution Writing the equation as

Example 2.1.69 Solve cos 2 x

[JNTU 1999S]

dy + sec2 x y = tan x sec2 x dx Here P ( x ) = sec2 x,

Q ( x ) = tan x sec2 x. This is linear in y. Also,

∫ P ( x) dx = ∫ sec

Integrating factor = e

2

x dx = tan x

tan x

The general solution is ye tan x = c + ∫ tan x sec2 xe tan x dx Put tan x = t ,

sec2 xdx = dt = c + ∫ t e t dt = c + (t − 1) e t

Integrating by parts y = ce − tan x + (tan x − 1). Example 2.1.70 Solve xy ′ + y + 4 = 0.

„ [JNTU 2001]

Solution Writing the equation in the form dy 1 4 + y=− , dx x x

Ch02.indd 61

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2-62

Differential Equations

we note that this is a linear equation in y with 1 4 , Q( x ) = − x x But, as it is, the equation can be written as P ( x) =

d ( xy ) + 4 dx = 0 which, on integration, yields the general solution xy + 4 x + c = 0

„

dy + 2 xy = 1. dx Solution This is a linear equation in y with 2x 1 P( x) = , Q( x ) = 2 x −1 x −1 But, as it is, the equation can be written as

Example 2.1.71 Solve ( x 2 − 1)

[JNTU 1999]

d [( x 2 − 1) y ] = 1 dx The general solution is

( x 2 − 1) y = x + c Example 2.1.72 Solve x

„

dy + y = log x. dx

[JNTU 1996S]

Solution This is a linear equation in y with P ( x) =

1 , x

Q ( x) =

log x x

But, as it is, it can be written as d ( xy ) = log x dx

∫ log x dx = x log x − x Integrating, the general solution is xy = c + ( x − 1) log x

Ch02.indd 62

„

12/9/2011 11:37:09 AM

Differential Equations of First Order and First Degree

Example 2.1.73 Solve x ( x − 1)

2-63

dy − y = x 2 ( x − 1)3 . dx

Solution The equation can be put in the standard form dy 1 − ⋅ y = x ( x − 1)2 dx x ( x − 1)

This is linear in y with P ( x) = −

Now,

1 , x ( x − 1)

⎛1

Q ( x ) = x ( x − 1)2

1 ⎞

∫ P ( x) dx = ∫ ⎜⎝ x − x − 1⎟⎠ dx

⎛ x ⎞ = log ⎜ ⎝ x − 1⎟⎠ Integrating factor = e ∫ P ( x ) dx = e log x ( x −1) x = x −1 The general solution is x ⎛ x ⎞ y⎜ dx = c + ∫ ( x − 1)2 x ⎝ x − 1⎟⎠ x −1 xy = c + ∫ ( x 3 − x 2 ) dx x −1 1 1 = c + x 4 − x3 . 4 3 Example 2.1.74 Solve

„

5x2 dy 2 . − y= (2 + x )(3 − 2 x ) dx x [JNTU 2004 (Set 3)]

Solution Divide by x 2 1 dy 2 5 − 3 y= 2 x dx x (2 + x ) (3 − 2 x )

Ch02.indd 63

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2-64

Differential Equations

2 ⎞ ⎛ 1 ⎞ ⎛ 1 y = 5∫ ⎜ dx + 2 ⎟ ⎠ ⎝ 2 + x 3 − 2 x ⎟⎠

⇒

∫ d ⎜⎝ x

⇒

y = 5 [log(2 + x ) − log (3 − 2 x )] + c. x2

Example 2.1.75 Solve ( x + 1)

„

dy − ny = e x ( x + 1) n+1. dx [JNTU 2004 (Set)]

Solution Divide by ( x + 1) n+1 ( x + 1) − n

dy − n( x + 1) n+1 y = e x dx

⇒ ⇒ Example 2.1.76 Solve cos x

∫ d[ x + 1)

−n

y ] = ∫ e x dx + c

( x + 1) − n y = c + e x dy + y sin x = 1. dx

dy + y sin x = 1 dx Eq. (2.82) can be written as dy + y tan x = sec x dx Here P = tan x; Q = sec x

Solution

„

cos x

(2.82)

(2.83)

∫ pdx = ∫ tan x dx = logsec x ∴

I.F. = e ∫ pdx = e logsec x = sec x

Multiplying (2.83) by sec x, we have

sec x ∴

Ch02.indd 64

dy + y sec x tan x = sec2 x dx

y sec x = ∫ sec2 xdx = tan x + c

(2.84) „

12/9/2011 11:37:10 AM

Differential Equations of First Order and First Degree

Example 2.1.77 Solve

2-65

dy 2 + 2 xy = e − x . dx

Solution

Here

dy 2 + 2 xy = e − x dx 2 p = 2 x, Q = e − x

∫ pdx = ∫ 2 xdx = x

(2.85) (2.86) 2

I.F. = e ∫ pdx = e x ∴ Multiplying Eq. (2.85) by e x2 , we get

2

dy 2 + 2 xye x = 1 dx or on integration, we obtain 2 ye x = x + c ex

2

(2.87)

(2.88) „

Example 2.1.78 Solve ( x + 2 y 3 )

dy = y. dx

Solution ( x + 2 y3 )

dy =y dx

Eq. (2.89) can be written as dx 1 − x = 2 y2 dy y

(2.89)

(2.90)

1 − ∫ dy 1 1 Eq. (2.90) is linear in x; Here p = − ,I.F. = e y = y y 1 Multiplying (2.90) by y we get

1 dx 1 − x = 2y y dy y 2 whose solution is ⎛ 1⎞ x ⎜ ⎟ = y2 + c ⎝ y⎠

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2-66

Differential Equations

EXERCISE 2.7 Solve the following: dy 1. x log x + y = x sin 2 x. dx 1 Ans: y log x = C − cos 2 x 2 dy 2. x cos x + y ( x sin x + cos x ) = 1. dx xy Ans: sec x = tan x + C 3.

[JNTU 2003, 2004]

dy + 2 y = e x (3sin 2 x + 2cos 2 x ). dx Ans: y = Ce −2 x + e x sin 2 x

4. y ′ + y tan x = sin 2 x, y(0) = 1 or

y = cos x − cos x + c cos x y(0) = 1 ⇒ 1 = c.

[Hint: y sec x = sin 2 x + c]

3

Ans: y = 2cos x − cos3 x 5. dx = ( x + y + 1) dy. Ans: x + y + 2 = ce y 6. y ′ + y cot x = 2 x cosec x. Ans: y = ( x 2 + c) cosec x 7. x

dy + 2 y = x 2 log x. dx

Ans: 16 x 2 y + x 4 (1 − 4 log x ) = c 8.

dy + y tan x = cos3 x. dx Ans: 4 y sec x = 2 x + sin 2 x + c

9. (1 − x 2 )

dy + 2 xy = x 1 − x 2 . dx

Ans: y = 1 − x 2 + (1 − x 2 )

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Differential Equations of First Order and First Degree

2-67

dy + (2 x 2 − 1) y = x 3 . dx Ans: y = x + cx 1 − x 2

10. x (1 − x 2 )

dx = sec y − x tan y. dy Ans: x sec y = tan y + c 12. ye y dx = ( y 3 + 2 xe y dy ).

11.

Ans: 13.

x + e− y = c y2

dy + y cot x = 4 x cos ec x y(π 2) = 0. dx Ans: y sin x = 2 x 2 −

π2

2 dy = y / (2 y log y + y − x ). 14. dx c Ans: x = + y log y y 15.

1 − y 2 dx = (sin −1 y − x ) dy. Ans: x = sin −1 y − 1 + ce − sin

16. sin x cos x

−1

y

dy + y = xe x cos 2 x. dx [Hint: tan x

dy + y sec2 x = xe x ] dx

on multiplication by sec2 x ⇒ ∫ d ( y tan x ) = ∫ xe x dx. Ans: y tan x = c + ( x − 1)e x dy + y = tan x. dx Ans: y = tan x – 1 + ce – tan x

17. cos 2 x

dy + y ( x sin x + cos x ) = 1. dx Ans: yx sec x = tan x + c

18. x cos x

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2-68

Differential Equations

19. x log x

dy + y = 2log x. dx

Ans: y log x = c + ( log x )

2

20. sin x cos x

dy = y + sin x. dx

Ans: y cot x = c + log tan 21.

22.

x 2

dy 1 − tan y = (1 + x ) e x sec y. dx 1 + x Ans: sin y = (1 + x ) ( e x + c ) dy 1 – 2 x + y = 1. dx x2 Ans: 2 x = e y (1 + x 2 )

23.

dy x 1 + y= . 2 dx 1 + x 2 x (1 + x 2 ) 1 ⎛1 ⎞ Ans: y 1 + x 2 = c + log tan ⎜ tan –1 x ⎟ ⎝2 ⎠ 2

24. (1 + y + x 2 y ) dx + ( x + x 2 ) dy = 0. Ans: xy = c – tan –1 x

2.1.7 Bernoulli’s Equation An equation of the form dy + Py = Qy a ( a ∈ R, a ≠ 1) (2.91) dx is called Bernoulli’s equation. Bernoulli’s equation is reducible to the linear form by the substitution z = y1− a . Multiplying Eq. (2.91) by (1 − a) y − a , it becomes (1 − a) y − a

Ch02.indd 68

dy + P (1 − a) y1− a = Q (1 − a) dx

(2.92)

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Differential Equations of First Order and First Degree

or P1 = (1 − a) P , z = y1− a ,

dz + P1z = Q1 dx Q1 = (1 − a)Q dz = (1 − a) y − a

2-69

(2.93)

which is linear in z. Its general solution is ze ∫

= y1− a e ∫

(1− a ) Pdx

(1− a ) Pdx

= c + ∫ (1 − a)Qe ∫

(1− a ) Pdx

(2.94)

Note 2.1.79 There are also equations that are not of the Bernoulli type shown in Eq. (2.91). These equations are reducible to linear form by appropriate substitution (See Example 2.1.92). dy − y tan x = − y 2 sec x. dx Solution This is Bernoulli’s type equation. Example 2.1.80 Solve

P = − tan x,

Q = − sec x,

a = 2.

Multiplying the equation by − y12 , we have − Put

1 dy 1 + (tan x ) = sec x 2 y dx y

1 1 = z ⇒ − 2 dy = dz y y dz + (tan x ) z = sec x dx

(This is a linear equation.) Integrating factor e ∫ Pdx = e ∫ tan x = e logsec x = sec x The general solution is z sec x =

Ch02.indd 69

1 sec x = c + ∫ sec2 xdx = c + tan x. y

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2-70

Differential Equations

Example 2.1.81 Solve y(2 xy + e x )dx = e x dy. Solution dy − y = 2 xe − x ⋅ y 2 dx This is Bernoulli’s equation. P = −1, Q = 2 xe − x , a=2 1 Multiplying the equation by − y2 , we have

1 y2 1 − 2 y −

dy + dx dy + dx

1 = −2 xe − x y 1 = −2 xe − x y

1 This is a linear equation in y = z Integrating factor = e ∫ 1dx = e x The general solution is ze x =

1 x e +c y

⇒ ∫ 2 xe − x e x dx = c − x 2 ⇒

ex = c − x2 . y

Example 2.1.82 Solve xy(1 + xy 2 )

„

dy = 1. dx

Solution dx − y ⋅ x = y3 x2 dy 1 This is linear in x . 1 Multiplying by − x2 , we have

−

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1 dx 1 + y ⋅ = − y3 2 x dy x

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Differential Equations of First Order and First Degree

Put

2-71

1 dz = z⇒ + yz = − y 3 x dy e∫

ydy

y2

+ =e ∫2

The general solution is y2

1 y22 e x y2 = c − ∫ y 3e 2 dy

ze 2 =

Put

y2 = u; 2

ydy = du = c − ∫ 2ue u du = c − 2(u − 1) e u

y2 1 y22 e = c − ( y 2 − 2) e 2 x y2 1 = ce 2 + 2 − y 2 . „ x dy + x sin 2 y = x 3 cos 2 y. Example 2.1.83 Solve dx Solution This not strictly in the form of Bernoulli’s equation (2.91). But we can write it as dy sec2 y + 2 x tan y = x 3 dx

Put

tan y = z ⇒ sec2 ydy = dz dz + 2 xz = x 3 , dx

which is a Bernoulli type Eq. (2.91). Integrating factor = exp ∫ 2 xdx = e x

2

ze x = tan ye x 2 = c + ∫ x 3e x dx 1 = c + ∫ te t dt 2 2

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2

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2-72

Differential Equations

x 2 = t ⇒ 2 xdx = dt

Put

1 = c + (t − 1) e t 2 The general solution is 1 2 2 tan y e x = c + ( x 2 − 1) e x 2 x2

tan y = ce − 2 +

or

1 2 ( x − 1). 2

„

EXERCISE 2.8 Solve the following: dy + y = x3 y6 . dx Ans: (5 + cx 2 ) x 3 y 5 = 2

1. x

2. xy (1 + xy 2 ) Ans: 3.

1 x

[JNTU 1995, 2002, 2004]

dy = 1. dx y2

= 2 − y 2 + ce − 2

dy y y + log y = (log y )2 . dx x x Ans: (log y ) −1 = 1 + cx

4. 2 xy

dy = ( x 2 + y 2 + 1). dx

Ans: y 2 − x 2 = cx − 1 5. x( x − y ) dy + y 2 dx = 0. Ans: 6. tan y

y x

= log y + c

dy + tan x = cos y cos 2 x. dx

Ans: cos y = cos x(sin x + c)

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Differential Equations of First Order and First Degree

2-73

dy + y 2 (1 − sin x ) cos x, y(0) = 2. dx Ans: 2 (tan x + sec x ) = y (2sin x + 1)

7. y = cos x

8. y (2 xy + e x ) dx = e x dy. Ans: e x = y(c − x 2 ) dy 9. tan y + tan x = cos y cos 2 x. dx Ans: sec y = (c + sin x ) cos x 10.

dy tan y − = (1 + x ) e x sec y. dx 1 + x Ans: sin y = (1 + x ) (e x + c)

dy + y = y 2 log x. dx 1 Ans: = log x + 1 + cx y dy y 12. + = x2 y3 . dx x 1 5 1 Ans: 5 5 = +c x y 2 x2 dy + y cot x = y 2 sin 2 x cos 2 x. 13. dx 1 1 Ans: = sin x cos3 x + c sin x y 3 11. x

14. y ( 2 xy + e x ) dx – e x dy = 0. Ans: e x + yx 2 + cy = 0 dy 2 2 − sin y 2 = ( x + 1) . 15. 2 y cos y 2 dx x + 1 sin y 2 1 3 = ( x + 1) + c Ans: x +1 3 dy 16. 3 x (1 – x 2 ) y 2 + ( 2 x 2 – 1) y 3 = ax 2 . dx y3 a = +c Ans: 2 x 1– x 1 – x2

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2-74

2.2

Differential Equations

APPLICATIONS OF ORDINARY DIFFERENTIAL EQUATIONS

Newton’s Law of Cooling Physical experiments show that the rate of change of temperature T with respect to time t , dT dt , of a body is proportional to the difference between the temperature of the body (T) and that of the surrounding medium (T0). This principle is known as Newton’s Law of Cooling and is expressed through the following first order and first degree differential equation: dT = − k (T − T0 ) ( k > 0) dt Separating the variables dT = − kdt (T − T0 )

(2.95)

(2.96)

Integrating, we get log(T − T0 ) = − kt + log c ⇒ (T − T0 ) = ce − kt

(2.97)

If Ti is the initial temperature of the body when t = 0, from Eq. (2.97) Ti − T0 = c (2.98) Eliminating c between Eqs. (2.97) and (2.98), we get (T − T0 ) = (Ti − T0 ) e − kt or

T (t ) = T0 + (Ti − T0 ) e − kt .

(2.99)

Method of solving the problem of Newton’s Law of Cooling 1. Identify T0, the temperature of the surrounding medium. Then the general solution is given by Eq. (2.99). 2. Use two given conditions and find the constant of integration c and the proportionality constant k. 3. Substitute c and k obtained from step 2 in Eq. (2.99). We can determine (i) the value of T for a given time t or (ii) the value of t for a given temperature T from Eq. (2.99).

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Differential Equations of First Order and First Degree

2-75

Law of Natural Growth or Decay If the rate of change of a quantity y at any time t is proportional to y, then dy αy (2.100) dt If k is the constant of proportionality, then the required differential equation is dy = ky (2.101) dt where k is a real constant. For growth, k > 0, the differential equation is dy = ky dt

( k > 0)

(2.102)

For decay, the differential equation is dy = − ky dt

( k > 0).

(2.103)

Example 2.2.1 The temperature of a body initially at 80°C reduces to 60°C in 12 min. If the temperature of the surrounding air is 30°C, find the temperature of the body after 24 min. Solution Let T be the temperature of the body at time t. By Newton’s Law of Cooling dT dT = − k (T − T0 ) ⇒ = − kdt dt T − T0 ⇒ log (T − T0 ) = − kt + log c ⇒ T − T0 = ce − kt

(2.104)

Temperature of the surrounding medium T0 = 30 T = T0 + ce − kt = 30 + ce − kt

(2.105)

Initial temperature T = Ti = 80 when t = 0 Ti − T0 = 80 − 30 = ce 0 ⇒ c = 50

Ch02.indd 75

(2.106)

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2-76

Differential Equations

T = T0 + ce − kt = 30 + 50e − kt

∴ When

t = 12, T = 60 ⇒ 60 = 30 + 50e − k ⋅12 ⇒k =

When

1 ⎛ 5⎞ log ⎜ ⎟ ⎝ 3⎠ 12

t = 24, T = 30 + 50e − 12 (log 3 )⋅24 5

1

2

⎛ 3⎞ = 30 + 50 ⎜ ⎟ ⎝ 5⎠ 9 = 30 + 50 × = 48 25

„

Example 2.2.2 A body is heated to 105°C and placed in air at 15°C. After 1 hr its temperature is 60°C. How much additional time is required for it to cool to 37 12 °C? Solution Let T be the temperature of the body at time t. By Newton’s Law of Cooling dT dT = − k (T − T0 ) ⇒ = − kdt dt T − T0 ⇒ log (T − T0 ) = − kt + log c ⇒ T − T0 = ce − kt Temperature of the surrounding medium T0 = 15 T = T0 + ce − kt = 15 + ce − kt Initial temperature (t = 0) T = Ti = 105 ⇒

When

t = 1, T = 60 ⇒ 60 = 15 + 90e 1 ⇒ e−k = 2 1 1 T = 37 2 ,37 2 = 15 + 90e − kt t

(2.109) − k ⋅1

t

22 12 ⎛ 1 ⎞ ⎛ 1⎞ = 15 + 90 ⎜ ⎟ ⇒ = ⎜ ⎟ ⇒t = 2 ⎝ 2⎠ 90 ⎝ 2 ⎠ Additional time required = 2 hr – 1 hr = 1 hr.

Ch02.indd 76

(2.108)

105 = 15 + ce 0

⇒ c = 90 When

(2.107)

„

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Differential Equations of First Order and First Degree

2-77

Example 2.2.3 The number N of bacteria in a culture grew at a rate proportional to N. The value of N was initially 100 and increased to 3 332 in 1 hr. What was the value of N after hrs? [JNTU 1996S, 2003] 2 Solution Here N is a natural number and is therefore discrete. But in view of its largeness N will be treated as a continuous variable which is a differentiable function of time t. The differential equation to be solved is dN dN = kN ⇒ = kdt dt N ⇒ log N = kt + log c ⇒ N = ce kt When

t = 0,

(2.110)

N = 100 ⇒100 = ce 0 ⇒ N = 100 e kt

t = 1hr = 3600sec, ⇒ e k ⋅1 =

N = 332 = log e kt

(2.111)

332 100

(2.112) 3

When ⇒

3 t = hr 2

⎛ 332 ⎞ 2 N =100 (e k ) = 100 ⎜ ⎝ 100 ⎟⎠ 3 2

N2 ⎛ 332 ⎞ =⎜ ⎟ 10000 ⎝ 100 ⎠

3

⇒ (10 N )2 = 3323

⇒

N = 604.9

„

Example 2.2.4 A radioactive substance disintegrates at a rate proportional to its mass. When the mass is 10 mg, the rate of disintegration is 0.051 mg per day. How long will it take for the mass of 10 mg to reduce to half. [JNTU 1995] Solution The governing differential equation is dm = − km ( k > 0) dt where m is the mass of the substance. Separating variables and integrating log m = kt + log c ⇒ m = ce − kt

Ch02.indd 77

(2.113)

(2.114)

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2-78

Differential Equations

When mass m = 10 mg dm = −0.051 dt dm negative sign is to be taken since is decreasing rate. dt −0.051 = − k × 10 ∴ ⇒k =

0.051 = 0.0051 10

(2.115)

Eq. (2.114) becomes m = ce − (0.0051)t

t=0

When

(2.116)

m = 10 mg 10 = ce − (0.0051)10 ⇒ c = 10

∴

(2.117)

m = 10e − (0.0051) t

∴ We have to find t when m = 5 mg ⇒e ⇒

5 = 10e − (0.0051) t = 2 ⇒ 0.0051t = ln 2 0.6931 t= = 135.9 days 0.0051

(0.0051) t

„

EXERCISE 2.9 1. A body initially at 80°C cools down to 60ºC in 20 min. The temperature of the air is 40°C. Find the temperature of the body after 40 min. Ans: 50°C 2. The air temperature is 20°C. A body cools from 140°C to 80°C in 20 min. How much time will it take to reach a temperature of 35°C? Ans: 60 min

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Differential Equations of First Order and First Degree

2-79

3. If the temperature of the air is 20°C and the temperature of a body drops from 100°C to 75°C in 10 min. what will be its temperature after 30 min? When will the temperature be 30°C? Ans: 46°C, 55.5 min 4. Uranium disintegrates at a rate proportional to the amount present at any instant. If m1 and m2 gms of uranium are present at time t1 and t2, respectively. Show that the half-life of uranium is [(t2 − t1 ) log 2] / log ( m1 / m2 ) . 5. The rate at which bacteria multiply is proportional to the instantaneous number present. If the original number doubles in 2 hrs, in how many hours will it triple? [JNTU 1987, 2000] Ans:

2log3 hrs = 3.17 hrs log 2

6. The rate of decay of radium varies as its mass at a given time. Given that half-life of radium is 1600 yrs, find out the percentage of the mass of radium it will disintegrate in 200 yrs. Ans: 8.3% approximately

2.2.1 Geometrical Applications Orthogonal trajectories of a family of curves Definition 2.2.5 A curve which cuts every member of a given family of curves at a right angle is called an orthogonal trajectory of the given family of curves. A family of curves which are orthogonal to themselves are called self-orthogonal. 1. In the electrical field, the paths along which the current flows are the orthogonal trajectories of the equipotential curves. 2. In fluid mechanics, the stream lines and the equipotential lines are orthogonal trajectories of one another. 3. In thermodynamics, the lines of heat flow are perpendicular to isothermal curves.

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2-80

Differential Equations

Method for finding orthogonal trajectories (1) Cartesian coordinates (2.118) Let f ( x, y, c ) = 0 represent the equation of a given family of curves with single parameter ‘c’. Differentiating Eq. (2.118) with respect to ‘x’ and eliminating ‘c’ from the equation thus obtained and Eq. (2.118) we obtain a differential equation of the form dy ⎞ ⎛ (2.119) φ ⎜ x, y, ⎟ = 0 ⎝ dx ⎠ for the given family of curves. Suppose there passes a curve of the given family and a member of the orthogonal trajectories through a point P (x, y). Let m1 be the slope of the curve of the given family and m2 the slope of an orthogonal trajectory. Since the curves cut at right angles, we have dy so that m1m2 = −1 and m1 = dx 1 dx m2 = − dy = − dy dx

( )

dx dy in Eq. (2.119) by − , we get the dy dx differential equation for the orthogonal trajectories as Therefore, if we replace

⎛ ⎝

φ ⎜ x, y, −

dx ⎞ =0 dy ⎟⎠

(2.120)

Integrating Eq. (2.120), we obtain the equation for the orthogonal trajectories. Example 2.2.6 Find the orthogonal trajectories of the rectangular hyperbolas x2 – y2 = c where c is a parameter. Solution Equation of the given curves (2.121) x2 − y2 = c Differentiating Eq. (2.121) with respect to x, we get the differential equation of the given curves as dy x= y (2.122) dx

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Differential Equations of First Order and First Degree

2-81

dy dx the differential equation for in Eq. (2.122) by − dx dy the orthogonal trajectories is obtained as Replacing

⎛ dx ⎞ dx x − y ⎜− ⎟ = 0⇒ x + y =0 dy ⎝ dy ⎠ Separating the variables, we get dx dy + =0 x y Integrating, the equation for orthogonal trajectories is log x + log y = log k or xy = k

(2.123)

(2.124) „

Example 2.2.7 Find the orthogonal trajectories of the family of parabolas through the origin and focii on the y-axis. Solution Equation of the family of parabolas x 2 = 4 ay Differentiating Eq. (2.125) with respect to x, dy 2 x = 4a dx

(2.125)

(2.126)

Eliminating ‘a’ between Eqs. (2.125) and (2.126), the differential equation for the given curves is x2

dy = 2 xy dx

or

x

dy = 2y dx

(2.127)

dx dy in Eq. (2.127) by − , the differential equation for dy dx orthogonal trajectories is Replacing

⎛ dx ⎞ x ⎜ − ⎟ = 2y ⎝ dy ⎠

or

2y

dy +x=0 dx

(2.128)

Separating the variables and integrating, we get the equation for orthogonal trajectories as 2 y2 + x2 = c where ‘c’ is a constant.

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2-82

Differential Equations

Example 2.2.8 Find the orthogonal trajectories of the family of semicubical parabolas ay 2 = x 3 where ‘a’ is a parameter. Solution The equation of the given curves is ay 2 = x 3 Differentiating Eq. (2.129) with respect to ‘x’, we get dy 2ay = 3x 2 dx

(2.129)

(2.130)

Eliminating ‘a’ between Eqs. (2.129) and (2.130), the differential equation is obtained as dy 2x = 3y (2.131) dx dy dx Replacing in Eq. (2.131) by − , the differential equation for dx dy the orthogonal trajectories is obtained as ⎛ dx ⎞ 3y = 2x ⎜ − ⎟ ⎝ dy ⎠

or

3y

dy + 2x = 0 dx

(2.132)

Separating the variables and integrating, we obtain the equation for orthogonal trajectories as 3 y2 − 2x2 = c where ‘c’ is an arbitrary constant.

„

Example 2.2.9 Show that the system of confocal ellipses x2 y2 + = 1( λ , parameter) is self-orthogonal. a2 + λ b2 + λ Solution The equation of the system of ellipses x2 y2 + =1 a2 + λ b2 + λ Differentiating Eq. (2.133) with respect to ‘x’ 2x 2 yy ′ + 2 =0 a +λ b +λ 2

Ch02.indd 82

(2.133)

(2.134)

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Differential Equations of First Order and First Degree

2-83

⇒ λ ( x + yy ′ ) + ( a 2 yy ′ + b 2 x ) = 0 dy ⎞ −( a 2 yy ′ + b 2 x ) ⎛ ⇒λ = ⎜⎝ y ′ = ⎟⎠ ( x + yy ′ ) dx a 2 yy ′ + b 2 x ( a 2 − b 2 ) x ⎫ = ⎪⎪ x + yy ′ x + yy ′ a 2 yy ′ + b 2 x ( a 2 − b 2 ) yy ′ ⎬ ⎪ b2 + λ = b2 − = x + yy ′ x + yy ′ ⎪⎭ a2 + λ = a2 −

(2.135)

Eliminating λ from Eq. (2.133) using Eq. (2.135) ( x + yy ′ ) x 2 ( x + yy ′ ) y 2 − =1 ( a 2 − b 2 ) x ( a 2 − b 2 ) yy ′ ⎛ y⎞ ⇒ ( x + yy ′ ) ⎜ x − ⎟ = a 2 − b 2 y′ ⎠ ⎝ 1 It is clear that if we replace y′ by − y ′ , the same equation is obtained. „ Example 2.2.10 Show that the system of confocal and coaxial parabolas y 2 = 4 a( x + a) is self-orthogonal. Solution The equation of the given parabolas y 2 = 4 a ( x + a)

(2.136)

Differentiating Eq. (2.136) with respect to ‘x’, we have dy ⎞ ⎛ 2 yy1 = 4 a ⎜ y1 = ⎟ ⎝ dx ⎠

(2.137)

Eliminating ‘a’ from Eqs. (2.136) and (2.137), we get yy ⎞ ⎛ y 2 = 2 yy1 ⎜ x + 1 ⎟ ⎝ 2 ⎠

or

y = 2 xy1 + yy12

(2.138)

1 Replacing y1 by − y in the above differential equation for the given curves, we obtain the same equation for the orthogonal trajectories.

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2-84

Differential Equations

(2) Polar coordinates Let

f ( x , θ , a) = 0

(2.139)

be the equation of the given system of curves in polar coordinates where ‘a’ is a parameter. Differentiating Eq. (2.139) with respect to ‘ θ ’ we get another equation. Between these two equations, we eliminate the parameter ‘a’ and obtain the differential equation for the given system of curves as dr ⎞ ⎛ (2.140) F ⎜ r,θ , ⎟ = 0 ⎝ dθ ⎠ dθ dr in Eq. (2.140) by − r 2 If we replace (which amounts to interdr dθ changing the role of the polar subnormal and the polar sub-tangent), we get the differential equation for the orthogonal trajectories as dθ ⎞ ⎛ F ⎜ r,θ , − r 2 ⎟ = 0 (2.141) ⎝ dr ⎠ Solving this differential equation, we get the equation for the orthogonal trajectories. The following solved examples illustrate the procedure. „

Example 2.2.11 Find the orthogonal trajectories of the family of cardioids r = a(1 − cos θ ) where ‘a’ is a parameter. Solution The equation of the given cardioids is r = a (1 − cos θ ) (2.142) Differentiating Eq. (2.142) with respect to ‘ θ ’, we have dr (2.143) = a sin θ dθ Eliminating ‘a’ between Eqs. (2.142) and (2.143), we get r 1 − cos θ = dr ( dθ ) sin θ 2sin 2 θ2 θ = = tan θ θ 2sin 2 cos 2 2 The differential equation for the given curves dr θ = r cot dθ 2

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Differential Equations of First Order and First Degree

2-85

dr − r 2 dθ we obtain from Eq. (2.144) the differenby dθ dr tial equation for orthogonal trajectories as Replacing

− r 2 dθ θ = r cot dr 2 ⇒

(2.144)

dr θ = − tan dθ r 2

Integrating we get

(2.145)

θ

log r = 2log cos + log 2c 2

θ

= c(1 + cos θ ) 2 which is the equation for the orthogonal trajectories. ⇒ r = 2c cos 2

(2.146) „

Example 2.2.12 Find the equation of the system of orthogonal trajectories of the family of curves r n sin nθ = a n where a n is a parameter. Solution The given curves are r n sin nθ = a n (2.147) By logarithmic differentiation of Eq. (2.147) with respect to ‘ θ ’ n dr n cos nθ + =0 r dθ sin nθ 1 dr (2.148) + cot nθ = 0 r dθ which is the differential equation for the given curves. dr dθ by − r 2 in Eq. (2.148), we get the differential Replacing dθ dr equation for the orthogonal system as ⇒

1 dθ cos nθ − ⋅ r2 + =0 r dr sin nθ ⇒

Ch02.indd 85

n sin nθ dr =n cos nθ r

(2.149)

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2-86

Differential Equations

By integrating, we obtain the equation for orthogonal trajectories as log r n = − log cos nθ + n log c

(2.150)

r n cos nθ = c n

⇒

(2.151) „

EXERCISE 2.10 Find the orthogonal trajectories of the family of curves given in Table 2.2. Table 2.2 S. No.

Curves

Parameter

Orthogonal trajectories

Cartesian form 1.

y 2 = 4 ax (parabolas)

a

2x 2 + y 2 = c (ellipses)

2.

x 2 + y 2 + 2 gx + c = 0 (coaxial circles)

g

x 2 + y 2 + 2 fy − c ′ = 0 (circles)

3.

x +y

a

x +y

4.

x 2 − y 2 = cx

c

y( y 2 + 3x 2 ) = a

a

r 2 = c exp ⎜

a

r (1 − cos θ ) = 2c

2 3

2 3

=a

2 3

4

(asteroids)

3

4 3

4

=c

3

Polar form

Ch02.indd 86

⎛ θ2 ⎞ ⎝ 2 ⎟⎠

5.

rθ = a

6.

r=

7.

r n = a n cos nθ

a

r n = c n sin nθ

8.

r n cos nθ = a n

a

r n sin nθ = c n

2a (1 + cos θ )

12/9/2011 11:37:16 AM

3 Linear Differential Equations with Constant Coefficients 3.1

INTRODUCTION

In the previous chapter, we have studied the linear differential equation of the first order dy + P ( x ) y = Q( x ) dx

(3.1)

It is called a homogeneous (or reduced) equation if Q ( x ) ≡ 0 and a non-homogeneous equation if Q ( x ) ≡/ 0. If we substitute y = u + v in Eq. (3.1), we have d (u + v ) + P ( x )(u + v ) = Q ( x ) dx ⎡ du ⎤ ⎡ dv ⎤ ⇒ ⎢ + P ( x )u ⎥ + ⎢ + P ( x ) v − Q ( x ) ⎥ = 0 ⎣ dx ⎦ ⎣ dx ⎦ which is satisfied if

Ch03.indd 1

du + P ( x )u = 0 dx

(3.2)

dv + P ( x )v = Q( x ) dx

(3.3)

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3-2

Differential Equations

This shows that y = u + v is the general solution (or complete solution) of Eq. (3.1) if u and v satisfy Eqs. (3.2) and (3.3), respectively, i.e., if u( x ) = ce − ∫ Pdx

(3.4)

v( x ) = e − ∫ Pdx ∫ Qe ∫ Pdx dx

(3.5)

The function u( x ) which contains an arbitrary constant c is called the complementary function (C.F.) and v( x ) which contains no such arbitrary constant is called the particular integral (P. I.) of Eq. (3.1). There is one arbitrary constant in the C.F., because the order of the differential equation is 1. Example 3.1.1 Solve dy + y = 2e x . dx

(3.6)

Solution Clearly u( x ) = ce − x satisfies the homogeneous equation dy + y = 0 derived from Eq. (3.6). dx The general (complete) solution of Eq. (3.6) is y = ce − x + e x Thus, the general solution y of a linear equation comprises of two parts: y = yc + y p where yc , which satisfies the homogeneous equation and contains as many arbitrary constants as the order of the differential equation, and y p , which satisfies the non-homogeneous equation and contains no arbitrary constants.

3.1.1

Linear Differential Equations of the Second Order

Second order linear differential equations have many applications in physics and engineering such as in motion of mechanical systems and fundamental electric circuits. They also lead to the study of higher functions such as Bessel’s, Legendre’s, and hypergeometric functions, which have wide applications. A second order differential equation is called linear if it can be written in the standard form

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Linear Differential Equations with Constant Coefficients

3-3

d2 y dy + P ( x) + Q ( x) y = R ( x) 2 dx dx

( D 2 + P ( x ) D + Q ( x )) y = R ( x )

or

(3.7)

d d2 , D 2 ≡ 2 are the differential operators. When applied dx dx on a differentiable function, ‘D’ does the operation of differentiation and so it must always be written on the left-hand side of the function upon which it is operating.

where D ≡

1. ‘D’ is a linear operator: D (c1 y1 ( x ) + c2 y2 ( x ) ) = c1 Dy1 ( x ) + c2 Dy2 ( x ) 2. ‘D’ satisfies index laws: D 2 = D ⋅ D, D 3 = D ⋅ D ⋅ D D ⋅ D n = D n D m = D m+ n ( D 0 = 1; , D 0 f ( x ) ) = 1⋅ f ( x) = f ( x) m

Note 3.1.2 Equation

dy + P ( x ) y = Q ( x ), in operator notation, is dx

( D + P ) y = Q ( x ) ⇔ y = e − ∫ Pdx ∫ Qe ∫ Pdx dx This gives a meaning to the operator D + P . A differential equation that cannot be written in the form Eq. (3.7) is called non-linear (see Examples 3.75 and 3.76) If R( x ) ≡ 0 then Eq. (3.7) becomes d2 y dy + P ( x) + Q ( x) y = 0 2 dx dx or

( D 2 + P ( x ) D + Q ( x )) y = 0

(3.8)

and is called a homogeneous (or reduced) equation (illustrative Examples (2) (3) and (5) below) and if R( x ) ≡/ 0 then Eq. (3.7) is called a non-homogeneous equation (Illustrative examples (1), (4) and (6) below.) Here P, Q, R are continuous functions of x on some interval I, which is to be understood even if we do not mention it in each case, hereafter.

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3-4

Differential Equations

1. y + 4 y = e − x sin x. 2. (1 − x 2 ) y ′′ − 2 xy ′ + 6 y = 0. 3. y ′′ + y ′ 2 = 0. 4. y ′′ = ( y ′ 2 + 1). 5. y ′′ − y = 0. 6. y ′′ + y ′ = tan x + sec2 x. A solution of Eq. (3.7) is a twice differential function y on some open interval I, which satisfies Eq. (3.7) identically. As in the case of the first order linear differential equation, the general solution of Eq. (3.7) consists of two parts yc and y p . If u( x ) and v( x ) satisfy Eqs. (3.8) and (3.7), respectively, u ′′( x ) + P ( x )u ′( x ) + Q ( x )u( x ) = 0 v ′′( x ) + P ( x )v ′( x ) + Q( x )v( x ) = R( x )

(3.9) (3.10)

then their sum y = u( x ) + v( x ) satisfies Eq. (3.7). u( x ) is called the complementary function (C.F.) and v( x ) is called the particular integral (P. I.) of Eq. (3.7). Example 3.1.3 Solve y ′′ + y ′ = 2e x .

(3.11)

Solution Clearly e x is a solution of the complete Eq. (3.11). Now consider the reduced equation y ′′ + y ′ = 0

(3.12)

1 and e − x are two solutions of Eq. (3.12). A linear combination of these solutions namely y = c1 + c2 e − x is also a solution of Eq. (3.12). The general solution of Eq. (3.12) is y = c1 + c2 e − x + e x

(3.13)

3.1.2 Homogeneous Equations—Superposition or Linearity Principle We begin our discussion of solution of a second order linear equation with the homogeneous equation. Let us consider an example.

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Linear Differential Equations with Constant Coefficients

3-5

Example 3.1.4 Solve y ′′ − y = 0

or

( D 2 − 1) y = 0.

(3.14)

Solution y1 = e x and y2 = e − x are solutions of the homogeneous linear differential equation for all x since ( D 2 − 1)e x = D 2 e x − 1 ⋅ e x = e x − e x = 0 ( D 2 − 1)e − x = D 2 e − x − 1 ⋅ e − x = ( −1)2 e − x − e − x = 0 In fact, a linear combination of y1 and y2 , i.e., y = c1 y1 + c2 y2 is also a solution of Eq. (3.14) since ( D 2 − 1) y = ( D 2 − 1)(c1 y1 + c2 y2 ) = ( D 2 − 1) (c1e x + c2 e − x ) = D 2 (c1e x + c2 e − x ) − (c1e x + c2 e − x )

= c1 ( D 2 e x ) + c2 ( D 2 e − x ) − (c1e x + c2 e − x ) = c1e x + c2 ( −1)2 e − x − c1e x − c2 e − x = 0

(3.15)

This is known as the superposition or linearity principle.

3.1.3

Fundamental Theorem for the Homogeneous Equation

Theorem 3.1.5 For a homogeneous linear differential Eq. (3.8), any linear combination of two solutions is again a solution. Proof Let y1 and y2 be solutions of Eq. (3.8). Then substituting y = c1 y1 + c2 y2 ( D 2 + P ( x ) D + Q( x )) y = ( D 2 + P ( x ) D + Q( x ))(c1 y1 + c2 y2 ) = D 2 (c1 y1 + c1 y2 ) + P ( x ) D (c1 y1 + c2 y2 ) + Q ( x )(c1 y1 + c2 y2 ) = c1 [ D 2 y1 + P ( x ) Dy1 + Q( x ) y1 ] + c2 [ D 2 y2 + P ( x ) y2 + Q( x ) y2 ] = c1 .0 + c2 .0 = 0 [∵ y1 , y2 are solutions of Eq. (3.15)] Note 3.1.6 The above principle does not hold for non-homogeneous equations and non-linear equations.

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3-6

Differential Equations

Non-homogeneous linear differential equation Consider the non-homogeneous differential equation ( D 2 − 1) y = 1 y1 = e x − 1 and y2 = e − x − 1

(3.16)

are solutions of Eq. (3.16) since ( D 2 − 1) y1 = ( D 2 − 1)(e x − 1) = D 2 (e x − 1) − 1(e x − 1) = ex − ex + 1 = 1 2 ( D − 1) y2 = ( D 2 − 1)(e − x − 1) = D 2 (e − x − 1) − 1(e − x − 1) = e− x − e− x + 1 = 1 But their sum y1 + y2 = (e x − 1) + (e − x − 1) = e x + e − x − 2 is not a solution. ( D 2 − 1)( y1 + y2 ) = D 2 (e x + e − x − 2) − 1(e x + e − x − 2) = e x + e − x − (e x + e − x − 2) ≠ 1 Also, 2 y1 = 2(e x − 1) = 2e x − 2 is not a solution ( D 2 − 1)(2 y1 ) = 2( D 2 − 1)(e x − 1) = 2[ D 2 (e x − 1) − 1(e x − 1)] = 2[e x − e x + 1] = 2 ≠ 1. Non-linear differential equation y1 = x 2 , y2 = 1 are solutions of the non-linear differential equation yy ′′ − xy ′ = 0

(3.17)

but ( −1) y2 = − x 2 and their sum y1 + y2 = x 2 + 1 are not solutions of Eq. (3.17).

3.1.4 Initial Value Problem (IVP) For a first-order differential equation, a general solution contains one arbitrary constant c. If an initial condition y( x0 ) = y0 is given, then a particular solution in which c will have a definite value is obtained. Example 3.1.7 Solve the initial value problem y′ +

Ch03.indd 6

1 1 y = 2 , y(1) = 0. x x

(3.18)

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Linear Differential Equations with Constant Coefficients

3-7

Solution This is a linear equation of the first order with 1 1 ;Q= 2; x x 1 ∫ Pdx = ∫ x dx = log x Integrating factor = e log x = x P=

Multiplying Eq. (3.18) by the integrating factor ‘x’ xy ′ + y =

1 1 ⇒ d ( xy ) = dx x x

Integrating and applying the initial condition xy = log x + c 1.0 = log1 + c ⇒ c = 0 The required solution is xy = log x For a second-order homogeneous linear Eq. (3.8), a general solution is of the form y = c1 y1 + c2 y2 involving two arbitrary constants c1 and c2 . An initial value problem now consists of Eq. (3.8) and two initial conditions y ( x0 ) = y0 ,

y ′ ( x0 ) = y0′

We have to find the particular solution satisfying these conditions as illustrated below. Example 3.1.8 Solve the initial value problem y ′′ + y = 0,

y(0) = 2,

y ′(0) = −1.

(3.19)

Solution cos x,sin x are solutions. We take y = c1 cos x + c2 sin x Differentiating with respect to x y ′ = − c1 sin x + c2 cos x

Ch03.indd 7

(3.20)

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3-8

Differential Equations

y (0) = c1 ⋅ 1 + c2 ⋅ 0 = 2 ⇒ c1 = 2 y ′(0) = −c1 ⋅ 0 + c2 ⋅ 1 = −1 ⇒ c2 = −1 Now the required particular solution is y = 2cos x − sin x

3.1.5

(3.21)

Linear Dependence and Linear Independence of Solutions

Two solutions y1 ( x ) and y2 ( x ) defined on some interval I are said to be linearly independent (L.I.) on I if ⇒

k1 y1 ( x ) + k2 y2 ( x ) = 0 k1 = k2 = 0

(3.22)

They are said to be linearly dependent (L.D.) if Eq. (3.22) holds, for some non-zero constants k1 and k2 . y1 and y2 are linearly independent if one is not proportional to the other. Criterion for linear independence of two functions y1 and y2 on I Two functions y1 and y2 defined on some interval I are linearly independent on I if the Wronskian (determinant) y y2 W ( y1 , y2 ) ( x ) = 1 ≠0 (3.23) y1′ y2′ Example 3.1.9 Show that y1 = cos x and y2 = sin x are linearly independent W (cos x,sin x ) =

cos x sin x = cos 2 x + sin 2 x = 1 ≠ 0. − sin x cos x

⎛π ⎞ Example 3.1.10 Show that y1 = cos(π + x ) and y2 = sin ⎜ + x ⎟ are ⎝2 ⎠ linearly dependent

W ( y1 , y2 ) =

Ch03.indd 8

cos(π + x ) sin ( π2 + x ) − cos x cos x = = 0. sin x − sin x − sin (π + x ) cos ( π2 + x )

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Linear Differential Equations with Constant Coefficients

3.1.6

3-9

General Solution, Basis and Particular Solution

A general solution of a second-order linear homogeneous (L.H.) equation y ′′ + P ( x ) y ′ + Q ( x ) y = 0 on an open interval I is of the form y = c1 y1 + c2 y2 where y1 and y2 are linearly independent on I (they are not proportional to each other) and c1 and c2 are arbitrary constants. Then, y1 and y2 are called the basis (fundamental system) of solution for the L.H. equation. A particular solution of L.H. equation is obtained by giving specific values to c1 and c2 in the general solution.

3.1.7

Second Order Linear Homogeneous Equations with Constant Coefficients

We will now discuss the method of solution of a homogeneous linear differential equation of the form ay ′′ + by ′ + cy = 0 or f ( D ) y = 0 where f ( D ) = aD 2 + bD + c

(3.24) (3.25)

To solve Eq. (3.24), we recall that a first-order linear equation y ′ − ay = ( D − a) y = 0 has an exponential function as solution: y = e ax . This gives us the idea to try as a solution of Eq. (3.24), the function y = e mx

(3.26)

We have Dy = me mx , D 2 y = m2 e mx . Substituting these in Eq. (3.24), we have ( am2 + bm + c)e mx = 0 ⇒ f ( m) = am2 + bm + c = 0 ∵ e mx ≠ 0

(3.27) (3.28)

Hence Eq. (3.25) is a solution of Eq. (3.26) if m is a solution of the algebraic Eq. (3.26) which is called the auxiliary equation or characteristic equation of Eq. (3.24).

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3-10

Differential Equations

It has, in general, two roots ⎛ − b + b 2 − 4 ac − b − b 2 − 4 ac ⎞ (α , β ) = ⎜ , ⎟ 2a 2a ⎝ ⎠

(3.29)

Note 3.1.11 Auxiliary equation f ( m) = am2 + bm + c = 0 is obtained from the homogeneous differential Eq. (3.24) by replacing ‘D’ by m. The general solution of Eq. (3.24) is obtained by combining two linearly independent solutions of Eq. (3.24) on I, which constitute a basis of solutions of Eq. (3.24) on I. Case (i): Two distinct real roots α , β . In this case, y1 = eα x , y2 = e β x constitute a basis of solutions of Eq. (3.24) on any interval I. Since these are linearly independent, the general solution is y = c1eα x + c2 e β x

(3.30)

Case (ii): Two real repeated (double) roots, α , α Let β = α . The solution is y = c1eα x + c2 eα x = (c1 + c2 ) eα x So, we have to find another independent solution for basis. We proceed as follows: Method 3.1.12 Let β = α + h. The solution is ⎡ ⎛ hx ( hx )2 y = c1eα x + c2 e ( α + h ) x = eα x ⎢c1 + c2 ⎜1 + + + ⎝ 1! 2! ⎣

⎞⎤ ⎟⎠ ⎥ ⎦

expanding e hx by exponential theorem. = eα x [(c1 + c2 ) + (c2 h) x ] assuming that terms with h2 and higher powers of h tend to zero. We can choose c2 to be sufficiently large so as to make hc2 finite as h → 0 and c1 large with a sign opposite to that of c2 so that c1 + c2 is finite. If c1 + c2 = A and hc2 = B, the general solution corresponding to two equal roots is y = ( A + Bx )eα x .

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Linear Differential Equations with Constant Coefficients

3-11

Method 3.1.13 If the auxiliary equation am2 + bm + c = 0 has roots equal to α am2 + bm + c = a ( m − α )2 The differential equation becomes ( D − α )2 y = 0 Put ( D − α ) y = z Then, ( D − α ) z = 0 dz (i.e.) − α z = 0 ⇒ z = Beα x dx Hence, ( D − α ) y = Beα x dy or − α y = Be − α x dx which is a linear equation and integrating factor = e − α x Its solution ye − α x = ∫ Beα x ⋅ e − α x dx = Bx + A ⇒ y = ( A + Bx ) eα x Case (iii): Since a, b, c are real, complex roots occur in conjugate pairs if the discriminant of the auxiliary equation f ( m) = am2 + bm + c = 0 is negative The general solution in this case is y = c1e ( α + iβ ) x + c2 e ( α −iβ x ) = eα x (c1e iβ x + c2 e − iβ x ) = eα x [c1 (cos β x + i sin β x ) + c2 (cos β x − i sin β x )] = eα x ( A cos β x + B sin β x ) where

A = c1 + c2 ; B = (c1 − c2 ) i are arbitrary constants.

d2 y dy + 3 + 2 y = 0. 2 dx dx Solution The given differential equation is ( D 2 + 3D + 2) y = 0

Example 3.1.14 Solve

The auxiliary equation is m2 + 3m + 2 = ( m + 1)( m + 2) = 0 ⇒ m = −1, −2 The general solution is y = c1e − x + c2 e −2 x .

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3-12

Differential Equations

Example 3.1.15 Solve

d2 y dy + 1.5 + 0.5 y = 0. dx 2 dx

Solution The given differential equation is (2 D 2 + 3D + 1) y = 0. The auxiliary equation is 2m2 + 3m + 1 = (2m + 1)( m + 1) = 0 ⇒ m = −1 2, −1 The general solution is y = c1e − x 2 + c2 e − x . Example 3.1.16 Solve (9 D 2 + 6 D + 1) y = 0. Solution The auxiliary equation is 9m2 + 6m + 1 = (3m + 1)2 = 0 ⇒ m = −1 3. − 1 3 The general solution is y = (c1 + c2 x ) e − x /3 . Example 3.1.17 Solve

d2x dx +k + k 2 x = 0. 2 dt dt

Solution The given differential equation is ( D 2 + kD + k 2 ) x = 0

d⎞ ⎛ ⎜⎝ D ≡ ⎟⎠ dt

The auxiliary equation is m2 + km + k 2 = 0 ⇒ m = −

k 3ki ± 2 2

The general solution is ⎛ 3kt 3kt ⎞ − kt 2 + c2 sin x = ⎜ c1 cos e 2 2 ⎟⎠ ⎝ d⎞ ⎛ Example 3.1.18 Solve ( D 2 + a 2 ) y = 0. ⎜ D = ⎟ ⎝ dx ⎠ Solution The auxiliary equation is m2 + a 2 = 0 ⇒ m = ± ia

The general solution is y = c1 cos ax + c2 sin ax

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Linear Differential Equations with Constant Coefficients

3-13

EXERCISE 3.1 Solve the following: 1. y ′′ + y ′ − 6 y = 0. Ans: y = c1e 2 x + c3e −3 x 2. 8 y ′′ − 2 y ′ − y = 0. Ans: y = c1e x 2 + c2 e − x 4 3. (9 D 2 + 12 D + 4) y = 0. Ans: y = (c1 + c2 x ) e −2 x 3 4. 4 y ′′ + 4 y ′ + y = 0. Ans: y = (c1 + c2 x ) e − x 2 5. y ′′ + π 2 y = 0. Ans: y = c1 cos π x + c2 sin π x 6. y ′′ + 2ky ′ + ( k 2 + 4 ) y = 0. Ans: y = e − kx (c1 cos 2 x + c2 sin 2 x ) 7.

d2 y dy + ( a + b) + aby = 0. 2 dx dx Ans: y = c1e − ax + c2 e − bx

8.

d2 y dy + 8 + 16 y = 0. 2 dx dx Ans: y = (c1 + c2 x ) e −4 x

9.

( D 2 − 3D + 4) y = 0.

(

Ans: y = e 3 x 2 c1 cos 7 x 2 + c2 sin 7 x 2

Ch03.indd 13

[JNTU 2003]

)

12/14/2011 11:32:53 AM

3-14

Differential Equations

( D 2 − 5D + 6) y = 0.

10.

Ans: y = c1e 2 x + c2 e 3 x

( D 2 − 6 D + 13) y = 0.

11.

⎛ 3 3 ⎞ −x 2 x + c2 sin x Ans: y = e ⎜ c1 cos 2 2 ⎟⎠ ⎝

( D 2 − 13D + 12) y = 0.

12.

Ans: y = c1e x + c2 e12 x

( D 2 + 4 ) y = 0.

13.

Ans: y = c1 cos 2 x + c2 sin 2 x 14. ( D 2 − D − 2) y = 0. Ans: y = c1e 2 x + c2 e − x 15. (3D 2 + D − 14 ) y = 0. Ans: y = c1e 2 x + c2 e −7 x 3 16. ( D 2 − 4 D + 1) y = 0.

(

Ans: y = c1e

3.1.8

3x

+ c2 e −

3x

)e

2x

Higher Order Linear Equations

The method of solution of second order linear equations can be extended to higher order equations. The differential equation P0

dn y d n −1 y + + P 1 dx n dx n −1

+ Pn y = X

(3.31)

is called an nth order differential equation with variable coefficients if at least one of the coefficients Pi (i = 0,1, , n) is a function of x and Eq. (3.29) is called an equation with constant coefficients if all Pi (i = 0,1, , n) are constants. The right-hand side member X in Eq. (3.29) is a function of x. Equation (3.29) is homogeneous if X ≡ 0 and non-homogeneous if X ≡/ 0 (Exercises 1, 3 and 4 below). In each term of Eq. (3.29), y and its derivatives occur in the first degree without being multiplied together.

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Linear Differential Equations with Constant Coefficients Table 3.1 S. No.

3-15

Classification of Linear Differential Equations

Equation

Type

Coefficients

1.

y′ v + y = e x

NH

constants

2.

y ′′′ − 6 y ′′ + 11 y − 6 y = 0

H

constants

3.

y ′′ +

NH

variables

4.

( x + 1) 2 y ′′ − 3( x + 1) y ′ + 4 y = x 2 + x + 1

NH

variables

5.

y ′′ −

y = 0 Bessel Equation

H

variables

⎛ ⎝

H

variables

H

variables

6.

1 x

7.

1 1 y ′ + 2 y = xe x x x

1

x2

( xy ′ ) + ⎜ 1 −

x2 ⎞

⎟

x2 ⎠

y = 0 Legendre’s Equation

[(1 − x 2 ) y ′ ]′ + x ( x + 1) y = 0

As in the case of linear equations of the first order, the general solution (or complete solution) of Eq. (3.29) comprises of two parts: y = yc + y p where y p , called the particular integral (P. I.) satisfies the complete non-homogeneous (N.H.) Eq. (3.29) and contains no arbitrary constants, while yc , called the complementary function (C.F.) (a linear combination of n linearly independent solutions containing n arbitrary constants) satisfies the homogeneous equation.

3.1.9

Linearly Independent (L.I.) Solutions

We now extend the definition of the two linearly independent functions given above to a set of linearly independent solutions of an nth order linear differential equation. A set of solutions { yi | i = 1,2, , n} defined on some interval I is said to be linearly independent (L.I.) on I if c1 y1 + c2 y2 + ⇒

Ch03.indd 15

c1 = c2 =

+ cn yn = 0 = cn = 0

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3-16

Differential Equations

It is linearly dependent (L.D.) on I if it is not linearly independent on I. The set of solutions { yi | i = 1,2, n} is linearly independent on I if the Wronskian determinant W ( y1 , y2 , y1 y1′ = y1( n −1)

, yn )( x0 ) y2 y2′ y2( n −1)

yn yn′

( x0 ) ≠ 0 ( x0 ∈ I )

yn( n −1)

The differential operator D We can conveniently denote dy d 2 y , , dx dx 2 as Dy, D 2 y, form

,

dn dx n

, D n y, respectively, and write Eq. (3.29) in the symbolic

f ( D ) y = ( P0 D n + P1 D n −1 +

+ Pn −1 D + Pn ) y = X

⎛ n ⎞ where f ( D ) = ⎜ ∑ Pr D n − r ⎟ y is a polynomial in the symbol D. ⎝ r=0 ⎠ When applied on a differentiable function, D does the operation of differentiation; so it must be always on the left-hand side of the function upon which we are applying it. It is a linear operator. D (c1 y1 + c2 y2 ) = c1 ( Dy1 ) + c2 ( Dy2 ) It satisfies index laws D m ⋅ D n = D n ⋅ D m = D m+ n D1 = D , D 2 = D ⋅ D , , D 0 = I [ D 0 f ( x ) = 1. f ( x ) = f ( x ) ].

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Linear Differential Equations with Constant Coefficients

3-17

3.1.10 Exponential Shift Theorem 3.1.19 If f ( D ) = D n + a1 D n −1 + a2 D n − 2 + are real constants, then

+ an where ai

eα x ( f ( D ) y ) = f ( D − α ) ( eα x y ) where y is a function of x. d αx ( e y ) − α eα x y dx dy = eα x + α eα x y − α eα x y = eα x ( Dy ) dx Apply ( D − α ) again

Proof ( D − α ) ( eα x y ) =

( D − α )2 ( eα x y ) = ( D − α ) ( eα x Dy ) = eα x D 2 y Repeating this r times, we have ( D − α ) r ( eα x y ) = eα x D r ⇒

n

r=0

⇒

n

n

r=0

r=0

∑ ar ( D − α )r (eα x y ) = ∑ ar (eα x D r y ) = eα x ∑ (ar D r ) y ( f ( D − α )) (e

αx

y ) = e f ( D) y αx

eα x f ( D ) y = f ( D − α ) ( eα x y )

or

This relation shows that the effect of shifting an exponential factor from the right-hand side of the operator to its left side is to replace D by ( D + α ).

Note 3.1.20 By the theorem on the exponential shift, we have f ( D − α )(eα xV ) = eα x ( f ( D )V )

(3.32)

where V is a function of x. Changing D to D + α in Eq. (3.32) we get f ( D )(eα xV ) = eα x ( f ( D + α )V )

(3.33)

Corollary 3.1.21 ( D − α ) n ( x k eα x ) = eα x D n ( x k ) 0 if ⎧ ⎪ αx =⎨ e n! if ⎪eα x k ( k − 1) ( k − n + 1) x k − n if ⎩

Ch03.indd 17

k n.

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3-18

Differential Equations

Corollary 3.1.22 n

Let φ ( D 2 ) = ∑ ar D 2 r r=0

φ ( D 2 ) sin α x = φ ( −α 2 ) sin α x

φ ( D 2 ) cos α x = φ ( −α 2 ) cos α x Rules 1. f ( D ) eα x = f (α ) eα x (Replace every D by α .) 2. f ( D 2 )

cos α x cos α x = f ( −α 2 ) sin α x sin α x (Replace every D 2 by − α 2 .)

(3.34)

Higher order Linear differential equation with constant coefficients We consider from now on, that the coefficients in Eq. (3.29) are all constants, i.e., Pi = ai constants. Method of finding the complementary function Consider the homogeneous equation f ( D) y = 0 or ( D n + a1 D n −1 +

+ an −1 D + an ) y = 0 ( a0 = 1)

If we put y = e mx as a trial solution we have the auxiliary equation f ( m) = mn + a1mn −1 +

+ an −1m + an = 0 [∵ e mx ≠ 0]

Thus, the solution of the homogeneous differential equation now reduces to that of an algebraic equation called the auxiliary equation obtained by replacing D with m. Solving the auxiliary equation we get n roots mi (i = 1,2 n), each giving rise to a solution. The general solution containing n arbitrary constants is obtained depending on the nature of roots in Table 3.2.

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Ch03.indd 19

eα x (c1 cos β x + c2 sin β x ) + c3e m3x + + cn e mn x eα x [(c1 + c2 x ) cos β x + (c3 + c4 x )sin β x ] + c5e m5x + + cne mn x

{e m1x , xe m1x , x 2e m1x , e m4 x ,… , e mn x }

{e m1x , xe m1x , x r −1e m1x , e mr +1x ,… e mn x }

{eα x cos β x, eα x sin β x, e m3x ,… e mn x } eα x cos β x, xeα x cos β x e ax sin β x, xeα x sin β x e m5x e mn x

Three equal, and (n − 3) distinct real roots m1, m1, m1, m4,…, mn

r equal, (n − r) distinct real roots m1, m1,…, m1(r roots) mr + 1,…, mn

Two complex conjugate roots (n − 2) distinct real roots, m1 = α + iβ , m2 = α − iβ

Two equal complex conjugate roots, (n − 2) distinct real roots, m1 = m2 = α + iβ , m3 = m4 = α − iβ

4.

5.

6.

2

+ cr +1e

1

2 mr +1x

3

2

m1x

+ c4 e m4 x + + cr x r −1 ) e m1x + + cne mn x

(c + c x +

1

(c + c x + c x ) e

+ cne mn x

+ cne mn x

3.

(c1 + c2 x ) e m1x + c3e m3x +

{e m1x , xe m1x , e m3x ,… , e mn x }

Two equal, and (n − 2) distinct real roots, m1, m1, m3, …, mn

+ cne mn x ]

2.

[c1e m1x + c2e m2 x +

n distinct, real roots m1, m2, …, mn

1.

{e m1x , e m2 x ,… , e mn x }

Roots

Case

General solution

Table 3.2 Linearly independent solutions

Linear Differential Equations with Constant Coefficients 3-19

12/14/2011 11:32:55 AM

3-20

Differential Equations

Example 3.1.23 Solve 4 y ′′′ + 4 y ′′ + y ′ = 0.

[JNTU 2003]

Solution The differential equation is (4 D 3 + 4 D 2 + D ) y = 0 The auxiliary equation is 1 1 4 m3 + 4 m2 + m = m (2m + 1)2 = 0 ⇒ m = 0, m = − , − 2 2 The general solution is y = (c1 + c2 x ) e − x 2 + c3 . Example 3.1.24 Solve ( D 3 + 27) y = 0. Solution The auxiliary equation is 3 3 3 m3 + 27 = ( m + 3) ( m2 − 3m + 9) = 0 ⇒ m = −3, ± 2 2 The general equation is ⎛ 3 3 3 3 ⎞ 3x 2 y = c1e −3 x + ⎜ c2 cos x + c3 sin x e . 2 2 ⎟⎠ ⎝ Example 3.1.25 Solve y ′′′ − y ′′ + 100 y ′ − 100 y = 0. Solution The auxiliary equation is m3 − m2 + 100 m − 100 = 0. Clearly m = 1 is a root. By synthetic division 1

1 1

–1 100 –100 1 0 100 0 100 0

∴ m3 − m2 + 100 m − 100 = ( m − 1) ( m2 + 10 2 ) = 0 ⇒ m = 1, ± 10i The general solution is y = c1e x + (c2 cos10 x + c3 sin10 x ).

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Linear Differential Equations with Constant Coefficients

3-21

d4 y + a 4 y = 0. dx 4 Solution The auxiliary equation is Example 3.1.26 Solve

m4 + a 4 = 0 m4 = − a 4 = a 4 e (2 k +1)π i m ⇒ = e (2 k +1)π i 4 k = 0,1, 2, 3 a ⎧ cos π4 + i sin π4 = 1+2i ⎪ 3π 3π −1+ i ⎪ cos 4 + i sin 4 = 2 =⎨ 5π 5π −1− i ⎪ cos 4 + i sin 4 = 2 ⎪ cos 7π + i sin 7π = 1− i 4 4 2 ⎩ The general solution is

(

)

(

)

⎡c cos ax 2 + c sin ax 2 ⎤ 2 ⎣ 1 ⎦ − ax 2 ⎡ +e c cos ax 2 + c4 sin ax 2 ⎤ . ⎣ 3 ⎦

y = e ax

2

(

)

(

)

Example 3.1.27 Solve (4 D 3 + 16 D 2 + 21D + 9) y = 0. Solution The auxiliary equation is 4 m3 + 16m2 + 21m + 9 = 0 ⇒ ( m + 1) (2m + 3)2 = 0 ⇒ m = −1, −3 2, −3 2 The general solution is y = c1e − x + (c2 + c3 x ) e −3 x 2 By synthetic division –1

4 4

16 21 9 –4 –12 –9 12 9 0

The general solution is y = c1e − x + (c2 + c3 x ) e −3 x 2

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3-22

Differential Equations

Example 3.1.28 Solve ( D 2 + 3) y = 0. 2

Solution The auxiliary equation is

( m2 + 3)2 = 0 ⇒ m = ±i

( repeated )

3

The general solution is y = (c1 + c2 x ) cos

( 3x ) + (c

3

+ c4 x )sin 3x.

EXERCISE 3.2 Solve: 1. ( D 3 + 3D 2 − 10 D ) y = 0. Ans: y = c1 + c2 e 2 x + c3e −5 x 2.

( D 2 − 4)2 y = 0. Ans: y = (c1 + c2 x ) e 2 x + (c3 + c4 x ) e −2 x ]

3.

d3x d2x dx − 2 − 3 = 0. 3 2 dt dt dt Ans: x (t ) = c1 + c2 e 3t + c3e − t ]

4.

( D 4 − 2 D 3 − 13D 2 + 38D − 24) y = 0. Ans: y = c1e x + c2 e 2 x + c3e 3 x + c4 e −4 x ]

5.

( D 3 + 3D 2 − 4) y = 0. Ans: y = c1e x + (c2 + c3 x ) e −2 x ]

6. ( 2 D 4 − 3D 3 − 2 D 2 ) y = 0. Ans: y = c1 + c2 x + c3e 2 x + c4 e − x 2 ]

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Linear Differential Equations with Constant Coefficients

7.

3-23

( D 2 + D + 1) y = 0.

(

)

(

)

Ans: y = e − x 2 ⎡c1 cos x 3 2 + c2 sin x 3 2 ⎤ ⎣ ⎦ 8.

( D 4 + 8D 2 + 16) y = 0. Ans: y = (c1 + c2 x ) cos 2 x + (c3 + c4 x )sin 2 x

9. ( D 3 + D 2 + 4 D + 4 ) y = 0 given that y = 0, y ′ = −1, y ′′ = 5 when x = 0. Ans: y = e − x − cos 2 x 10.

( D 3 + 6 D 2 + 3D − 10) y = 0. Ans: y = c1e x + c2 e −2 x + c3e −5 x

11.

( D 3 − 4 D 2 + 5D − 2) y = 0. Ans: y = (c1 + c2 x ) e x + c3e 2 x

12. ( D 4 + 2 D 3 − 3D 2 − 4 D + 4 ) y = 0. Ans: y = (c1 + c2 x ) e x + (c3 + c4 x ) e −2 x 13.

( D 4 − D 3 − 9 D 2 − 11D − 4) y = 0. Ans: y = (c1 + c2 x + c3 x 2 ) e − x + c4 e 4 x

14. ( D 4 − k 4 ) y = 0. Ans: y = c1e kx + c2 e − kx + c3 cos kx + c4 sin kx 15. ( D 4 + 8 D 2 + 16) y = 0. Ans: y = (c1 + c2 x ) cos 2 x + (c3 + c4 x )sin 2 x 16.

( D 4 + 3D 2 − 4) y = 0. Ans: y = c1e x + c2 e − x + c3 cos 2 x + c4 sin 2 x

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3-24

3.1.11

Differential Equations

1 Inverse Operator D -1 or D

Method of finding the particular integral If Q is a differentiable function of x on an interval I and D ≡

d is a dx

1 dQ , D −1 or is then called the D dx 1 ⎛1 ⎞ inverse operator such that ( DQ ) = D ⎜ Q ⎟ = 1 ⋅ Q = Q. ⎝D ⎠ D differential operator such that DQ =

1 Q = ∫ Qdx is the integral of Q ( x ) D If Q is a continuous function on an interval I, then n 1 Q , where f ( D ) = ∑ ar D n − r f ( D) r=0

is a function of x containing no arbitrary constant such that ⎛ 1 ⎞ f ( D) ⎜ Q ⎟ = Q. ⎝ f ( D) ⎠ Theorem 3.1.29 If Q is a continuous function of x on I and α is a real or complex constant, then a particular value of 1 Q is eα x ∫ Qe − α x dx D −α Proof Let y =

1 Q D −α

Operating ( D − α ) on both sides, we get ⎛ 1 ⎞ =Q (D − α) y = (D − α) ⎜ ⎝ D − α ⎟⎠

or

dy −αy = Q dx

which is a linear equation of the first order, whose solution is ye − α x = c + ∫ Qe − α x dx

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Linear Differential Equations with Constant Coefficients

3-25

1 Since we are interested here in a particular solution Q , we D −α take c = 0. Hence y = eα x ∫ Qe − α x dx Thus,

1 Q = eα x ∫ Qe − α x dx D −α

(3.35)

1 1 and are two inverse operators where α , β are given D−β D −α real/complex constants, then If

1 1 Q= D−β ( D − β )( D − α ) 1 = D−β

⎛ 1 ⎞ Q⎟ ⎜⎝ D −α ⎠

(e

αx

∫ Qe α dx ) = eβ ∫ ⎡⎣eα ∫ Qe α dx ⎤⎦ dx − x

x

x

− x

(3.36) This can be extended to n factors. The following method of putting 1 into partial fractions is of practical importance. f ( D)

3.1.12

General Method for Finding the P. I.

Theorem 3.1.30 If f ( D ) = ( D − α1 )( D − α 2 ) ( D − α n ) where α r are real or complex constants, then particular integral 1 1 Q= Q f ( D) ( D − α1 )( D − α 2 ) ( D − α n ) n ⎛ A1 An ⎞ A2 ≡⎜ + + + = Q Ar eα n x ∫ Qe − α r x dx ∑ D − α n ⎟⎠ ⎝ D − α1 D − α 2 r =1 =

where Ar is the determinable constant corresponding to the partial 1 fraction . D − αr

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3-26

Differential Equations

Example 3.1.31 Find particular values of 1 1 1 (ii) 2 e ax (iii) 3 cos ax. (i) x 3 D D D Solution 1 3 1 x = ∫ x 3 dx = x 4 . (i) D 4 1 ax 1 ⎛ 1 ax ⎞ 1 ⎛ e ax ⎞ 1 ⎛ 1 ax ⎞ 1 ax e = ⎜ e ⎟= ⎜ ⎟= ⎜ e ⎟= 2e . (ii) D2 D⎝ D ⎠ D⎝ a ⎠ a⎝ D ⎠ a (iii)

(

)

1 1 1 ⎛ sin ax ⎞ 1 1 ⎛ 1 ⎞ cos ax = 2 ∫ cos axdx = 2 ⎜ ⎟⎠ = ⎜⎝ sin ax ⎟⎠ 3 ⎝ D D D a aD D 1⎛ 1 ⎞ 1 1 ⎛ − cos ax ⎞ −1 cos axdx = ⎜ ∫ sin axdx ⎟ = ⎟= ⎠ a D ⎜⎝ a⎝ D a ⎠ a2 ∫ 1 sin ax 1 =− 2 = − 3 sin ax. a a a

Example 3.1.32 Find the particular values of 1 1 1 e2 x x (i) (ii) (iii) cos 2 x. D−2 D+2 D+5 Solution 1 1 x= x = e −2 x ∫ xe 2 x dx (i) D+2 D − ( −2) e 2 x ⎞ −2 x ⎛ x 1 ⎞ 2 x 1 ⎛ e2 x = e −2 x ⎜ x − ∫1 dx⎟ e ⎜ − ⎟ e = (2 x − 1). ⎝ 2 4⎠ ⎝ 2 ⎠ 2 4 1 e 2 x = e 2 x ∫ e 2 x ⋅ e −2 x dx = xe 2 x (ii) D−2 1 1 (iii) cos 2 x = cos 2 x = e −3 x ∫ cos 2 xe 3 x dx D+3 D − ( −3) e3 x = e −3 x 2 (3cos 2 x + 2sin 2 x ) 3 + 22 3cos 2 x + 2sin 2 x = . 13 Example 3.1.33 Find the particular value of 1 1 x. e2 x (ii) 2 (i) D −1 ( D − 1)( D − 2)

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Linear Differential Equations with Constant Coefficients

3-27

Solution 1 ⎛ 1 1 1 ⎞ e2 x ⎟ = e 2 x ∫ e 2 x e −2 x dx = (i) ( xe2 x ) ⎜⎝ D −1 D − 2 ⎠ D −1 D −1 = e x ∫ xe 2 x ⋅ e − x dx = e x ∫ xe x dx

(

(

)

) ⎛ 1 dx ) = e ⎜ − xe D +1 ⎝

= e x xe x − ∫ 1.e x dx = e x ( x − 1) e x = ( x − 1) e 2 x .

(ii)

(

1 1 e− x ⎞ x −x + x= e x ∫ xe − x ( D + 1) ( D − 1) ( −1) ⎟⎠ D +1 1 =− ( x + 1) = − e − x ∫ ( x + 1) e x dx D +1 = − e − x [( x + 1) e x − 1 ⋅ e x ] = − x

Alternative method (i)

(ii)

1 1 ⎞ 2x 1 1 2x ⎛ 1 e2 x = ⎜ e2 x − e − ⎟⎠ e = ⎝ D − 2 D −1 D−2 D −1 ( D − 1) ( D − 2) = e 2 x ∫ e 2 x ⋅ e −2 x dx − e x ∫ e 2 x e − x dx = xe 2 x − e x ⋅ e x = e 2 x ( x − 1). 1 x= ( D + 1) ( D − 1) = =

1⎛ 1 1 ⎞ − ⎜⎝ ⎟x 2 D − 1 D + 1⎠ 1 x 1 ⋅ e ∫ xe − x dx − e − x ∫ xe x dx 2 2 e− x ⎞ 1 x ⎛ e− x − ∫ 1⋅ e ⎜x dx 2 ⎝ ( −1) ( −1) ⎟⎠ 1 − e − x xe x − ∫ 1e x dx 2 1 x 1 e ( − x − 1) e − x − e − x ⋅ ( x − 1) e x 2 2 1 1 ( − x − 1) − ( x − 1) = − x 2 2

(

= =

)

EXERCISE 3.3 Find the particular value of each of the following: 1 1. 2 e 3 x . D e3 x Ans: 9

Ch03.indd 27

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3-28

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

Ch03.indd 28

Differential Equations

1 3 ( x + 1) . D2 x5 x2 Ans: + 20 2 1 [Hint : cos3 x = 14 (cos3 x + cos x )] cos3 x. D 1 ⎛ sin 3 x ⎞ + sin x ⎟ Ans: ⎜ ⎠ 4⎝ 3 1 ( e 3 x − 2) . 2 D −2 1 Ans: e 3 x + 1 7 1 cos3 x. D2 − 9 −1 cos3 x Ans: 18 1 x. 2 D −4 −x Ans: 4 1 2 3x . D Ans: x 3 1 2x e . D2 1 Ans: e 2 x 2 1 sin x . D2 Ans: − sin x 1 2sin 2 x. D 1 Ans: x − sin 2 x 2 1 x e. D +1 1 Ans: e x 2

12/14/2011 11:32:59 AM

Linear Differential Equations with Constant Coefficients

12.

3-29

1 x. D −1 Ans: x + 1

13.

1 e2 x. D −1 2

Ans: 14.

1 2x e 3

1 e2 x. D −4 2

Ans:

1 2x xe 4

1 4 x. D2 1 6 x Ans: 30 1 e ax. 16. ( D − a) 2 15.

Ans:

3.2

1 2 ax xe 2

GENERAL SOLUTION OF LINEAR EQUATION f ( D ) y = Q( x )

We have seen that if y = y p is a particular solution (containing no arbitrary constants) of the linear equation f ( D ) y = Q ( x ) and y = yc is the general solution of the homogenous equation f ( D ) y = 0 with as many arbitrary constants as the order of the equation then y = yc + y p is the general solution (G.S.) of the non-homogenous linear equation: f ( D ) y = Q( x ) As we have studied methods of finding yc and the general method of finding yp earlier, we will work out some examples and give some exercises.

Ch03.indd 29

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3-30

Differential Equations

The method of finding yp discussed above is particularly useful when Q ( x ) = tan ax,cot ax,sec ax or csc ax and also when we cannot use special short methods (discussed in section 3.2.1) when Q ( x ) = e ax , sin ax,cos ax, x m , e axV and xV . Example 3.2.1 Solve ( D 2 − 7 D + 6) y = e 2 x (1 + x ).

[JNTU 2003]

Solution To find the complementary function, we have to solve ( D 2 − 7 D + 6) y = 0 The auxiliary equation is m2 − 7m + 6 = ( m − 1) ( m − 6) = 0 ⇒ m = 1, 6 yc = c1e x + c2 e 6 x To find the particular integral, 1 1⎛ 1 1 ⎞ 2x e 2 x (1 + x ) = ⎜ − ⎟ (1 + x ) e D − 7D + 6 5 ⎝ D − 6 D − 1⎠ 1⎡ 1 1 ⎤ 1 = ⎢ (1 + x ) e 2 x − (1 + x ) e 2 x ⎥ = ( I1 − I 2 ) D −1 5⎣D −6 ⎦ 5 1 I1 = (1 + x ) e 2 x = e6 x ∫ (1 + x ) e 2 x .e −6 x dx = e 6 x ∫ (1 + x ) e −4 x dx D −6 ⎡ e −4 x e −4 x ⎤ e2 x = e6 x ⎢(1 + x ) −1 = − (5 + 4 x ) ⎥ ( −4) ( −4)2 ⎦ 16 ⎣ 1 I2 = (1 + x ) e 2 x = e x ∫ (1 + x ) e 2 x ⋅ e − x dx D −1 = e x ∫ (1 + x ) e x dx = e x [(1 + x ) e x − 1 ⋅ e x ] = xe 2 x

yp =

2

1 ⎡ e2 x e2 x ⎤ y p = ⎢− (5 + 4 x ) − xe 2 x ⎥ = − (1 + 4 x ) 5 ⎣ 16 16 ⎦ The general solution is y = yc + y p = c1e x + c2 e 6 x −

e2 x (1 + 4 x ). 16

Example 3.2.2 Solve ( D 3 + 2 D 2 − 3D ) y = xe 3 x . Solution To find the complementary function, we have to solve ( D 3 + 2 D 2 − 3D ) y = 0

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Linear Differential Equations with Constant Coefficients

3-31

The auxiliary equation is m3 + 2m2 − 3m = m ( m2 + 2m − 3) = m ( m − 1) ( m + 3) = 0 ⇒ m = 0,1, −3 yc = c1 + c2 e x + c3e −3 x To find the particular integral, yp =

1 1 xe 3 x = xe 3 x . f ( D) D ( D − 1) ( D + 3)

Partial fractions 1 B C ⎤ 3x ⎡A + xe 3 x = ⎢ + xe f ( D) ⎣ D D − 1 D + 3 ⎥⎦ ⎛1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ = A ⋅ ⎜ xe 3 x ⎟ + B ⎜ xe 3 x ⎟ + C ⎜ xe 3 x ⎟ ⎝D ⎠ ⎝ D −1 ⎠ ⎝ D+3 ⎠ 1 e3 x 1 e3 x 1 e3 x = − (3 x − 1) + ⋅ (2 x − 1) + (6 x − 1) 3 9 4 4 12 36 1 D =− A= 3 f ( D) D=0 B=

1 D −1 = f ( D ) D =1 4

1 D+3 = f ( D ) D = −3 12 e3 x 1 3x e3 x e3 x − 1 2 = (3 x − 1) xe = ∫ xe 3 x dx = x 9 3 3 D 1 e2 x e2 x ⎞ e3 x 3x x 3x − x x ⎛ − 2 ⎟ = (2 x − 1) xe = e ∫ xe e dx = e ⎜ x ⎝ 2 2 ⎠ 4 D −1 6x 6x 1 1.e ⎞ e3 x ⎛ e − 2 ⎟ = (6 x − 1) xe3 x = e −3 x ∫ xe 3 x e 3 x dx = e −3 x ⎜ x ⎝ 6 6 ⎠ 36 D+3 − 1 1 1 1 1 1 ( 1) x ⎡ ⎤ ⎛ ⎞ 3x = e3 x ⎢ x ⎜ − + + ⎟ + − − ⎥ = 36 e ⎝ ⎠ 9 8 72 27 6 432 ⎣ ⎦ C=

The general solution is y = c1 + c2 e x + c3e −3 x +

Ch03.indd 31

x − 1 3x e . 36

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3-32

Differential Equations

Example 3.2.3 Solve ( D 2 + a 2 ) y = sec ax.

[JNTU 2000]

Solution To find the complementary function, we have to solve ( D 2 + a2 ) y = 0 The auxiliary equation is m2 + a 2 = 0 ⇒ m = ± ia yc = c1 cos ax + c2 sin ax To find the particular integral, 1 sec ax ( D + ia) ( D − ia) 1 ⎛ 1 1 ⎞ 1 = − (I − I ) ⎜⎝ ⎟⎠ sec ax = 2ia D − ai D − ia 2ia 1 2 1 I1 = sec ax = e iax ∫ e − iax sec ax dx D − ai i cos ax − i sin ax ⎛ ⎞ dx = e iax ⎜ x + log cos ax ⎟ = e iax ∫ ⎝ ⎠ a cos ax i ⎛ ⎞ I 2 = e iax ⎜ x − log cos ax⎟ , ⎝ ⎠ a

yp =

Replacing i by −i in the above result, we have 1 ⎡ iax i ⎤ x ( e − e − iax ) + log cos ax ( e iax + e − iax )⎥ ⎢ 2ia ⎣ a ⎦ x 1 = sin ax + 2 cos ax log(cos ax ) a a

∴ yp =

The general solution is x 1 y = yc + y p = c1 cos ax + c2 sin ax + sin ax + 2 cos ax log cos ax. a a d2 y + a 2 y = tan ax. dx 2 Solution To find the complementary function, we have to solve Example 3.2.4 Solve

( D 2 + a2 ) y = 0

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Linear Differential Equations with Constant Coefficients

3-33

The auxiliary equation is m2 + a 2 = 0 ⇒ m = ± ia y = c1 cos ax + c2 sin ax To find the particular integral, 1 tan ax D + a2 1 ⎛ 1 1 ⎞ 1 yp = − (I − I ) ⎜⎝ ⎟⎠ tan ax = 2ia D − ia D + ia 2ia 1 2 1 I1 = tan ax = eiax ∫ eiax tan ax dx D − ia ⎛ cos ax − i sin ax ⎞ = eiax ∫ ⎜ ⎟⎠ sin ax dx ⎝ cos ax ⎡⎛ − cos ax ⎞ ⎤ = eiax ⎢⎜ − i ∫ (sec ax − cos ax ) dx ⎥ ⎟ a ⎠ ⎣⎝ ⎦ ∵ sin 2 ax = 1 − cos 2 ax ⎡ − cos ax i ⎤ = eiax ⎢ − {log(sec ax + tan ax ) − sin ax}⎥ a ⎣ a ⎦ e iax =− [(cos ax − i sin ax ) + i log(sec ax + tan ax ) a 1 I1 = − [1 + ieiax log (sec ax + tan ax ) ] a yp =

2

Replacing i by −i in the above result, we have 1 [1 − ieiax log (sec ax + tan ax)] a 1 y p = − 2 [1 + ieiax log (sec ax + tan ax ) ] − 1 2ia + ie iax log (sec ax + tan ax ) 1 = − 2 cos ax log (sec ax + tan ax ) a I2 = −

The general solution is y = yc + y p = c1 cos ax + c2 sin ax −

Ch03.indd 33

1 cos ax log (sec ax + tan ax ). a2

12/14/2011 11:33:01 AM

3-34

Differential Equations

Example 3.2.5 Solve ( D 2 − 1) y = 2e x + 3 x.

[JNTU 1996S]

Solution To find the complementary function, we have to solve ( D 2 − 1) y = 0 The auxiliary equation is m2 − 1 = 0 ⇒ m = ±1 yc = c1e x + c2 e − x To find the particular integral, 1⎛ 1 1 ⎞ x 1⎛ 1 1 ⎞ yp = ⎜ − − 2e + ⎜ ⎟ ⎟ 3x 2 ⎝ D − 1 D + 1⎠ 2 ⎝ D − 1 D + 1⎠ 3 = e x ∫ e − x e x dx − e − x ∫ e x e x dx + e x ∫ xe − x dx − e − x ∫ xe x dx 2 2x −x ⎛ e e e− x ⎞ 3 = xe x − e − x ⋅ + ex ⎜ x − 2 2 ⎝ ( −1) ( −1)2 ⎟⎠ 3 1 ⎛ ex ⎞ − e − x ⎜ x − e x ⎟ = xe x − e x − 3 x ⎝ 1 ⎠ 2 2

(

)

The general solution is 1 y = yc + y p = c1e x + c2 e − x + xe x − e x − 3 x. 2 Example 3.2.6 Solve ( D 2 + 1) y = cosec x. Solution To find the complementary function, we have to solve ( D 2 + 1) y = 0 The auxiliary equation is m2 + 1 = 0 ⇒ m = ± i yc = c1 cos x + c2 sin x To find the particular integral, 1⎛ 1 1 ⎞ − ⎜⎝ ⎟ cosec x 2i D − i D − i ⎠ 1⎛ 1 1 ⎞ = ⎜ cosec x − cosec x ⎟ ⎠ D+i 2i ⎝ D − i

yp =

Ch03.indd 34

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Linear Differential Equations with Constant Coefficients

3-35

1 ⎛ cos x − i sin x ⎞ cosec x = e ix ∫ e − x cosec x dx = eix ∫ ⎜ ⎟⎠ dx ⎝ D −i sin x ⎡ cos x ⎤ = e ix ⎢ ∫ dx − i ∫ dx ⎥ = e ix ( logsin x − ix ) ⎣ sin x ⎦ 1 cosec x = e − ix ( logsin x + ix ) D+i Replacing i by −i in the above result, we have yp =

e ix − e − ix e ix + e − ix logsin x − x = sin x logsin x − x cos x 2i 2

The general solution is y = yc + y p = c1 cos x + c2 sin x + sin x logsin x − x cos x. EXERCISE 2.4 Solve: 1. ( D 2 − 3D + 2) y = e 3 x . Ans: y = c1e x + c2 e 2 x +

e3 x 2

2. ( D 2 − 1) y = 3 + 7 x. Ans: y = c1 cosh x + c2 sinh x − 3 − 7 x 3. ( D 2 + 9) y = sec3 x. x cos3 x log cos3 x Ans: y = c1 cos3 x + c2 sin 3 x + sin 3 x + 3 9 4. ( D 2 + a 2 ) y = cos bx. Ans: y = c1 cos ax + c2 sin ax +

cos bx , a2 ≠ b2 a2 − b2

5. ( D 2 + 3D + 2) y = x 2 . 1⎛ 7⎞ Ans: y = c1e − x + c2 e −2 x + ⎜ x 2 − 3 x + ⎟ 2⎝ 2⎠

Ch03.indd 35

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3-36

Differential Equations

3.2.1 Short Methods for Finding the Particular Integrals in Special Cases ⎛ n ⎞ Let f ( D ) y = ⎜ ∑ ar D n − r ⎟ y = Q ( x ) ⎝ r=0 ⎠

(3.37)

be a non-homogenous linear equation, where ar are real constants. We will now consider short methods for finding the P.Is in special cases when Q(x) is of the form (1) e aX (2) sin ax or cos ax (3) x m (4) e aX V and (5) xV , where V is a function of x. (1) Q(x) = eax where ‘a’ is a constant Case (i) f ( a) ≠ 0 Since De ax = ae ax , D 2 e ax = a 2 e ax , and generally, D r e ax = a r e ax ⎛ n ⎞ ⎛ n ⎞ f ( D ) = e ax = ⎜ ∑ ar D n − r ⎟ e ax = ⎜ ∑ ar a n − r ⎟ e ax = f ( a) e ax ⎝ r=0 ⎠ ⎝ r=0 ⎠ Applying

(3.38)

1 on both sides of Eq. (3.35) f ( D)

⎛ 1 ⎞ 1 1 f ( D ) e ax = e ax ⎟ ( f ( a) e ax ) = f ( a) ⎜ f ( D) f ( D) ⎝ f ( D) ⎠ ⎛ 1 ⎞ 1 1 ax (3.39) ⇒ e ax = f ( a) ⎜ e ax ⎟ ⇒ e ax = e , f ( D) f ( a) ⎝ f ( D) ⎠ provided f ( a) ≠ 0

Rule Replace D in f ( D ) with a if f ( a) ≠ 0, i.e., ( D − a) is not a factor of f ( D ). Case (ii) f ( a) = 0, f ′( a) ≠ 0, i.e., ( D − a) is a factor of f ( D ) but ( D − a)2 is not a factor of f ( D ). Let f ( D ) = ( D − a)φ ( D ), where ( D − a) is not a factor of φ ( D ). We have f ′( D ) = φ ( D ) + ( D − a)φ ′( D ) ⇒ f ′( a) = φ ( a) 1 1 1 ⎛ 1 ⎞ e ax = e ax = e ax ⎟ , φ ( a) ≠ 0 ⎜⎝ f ( D) φ ( a) D − a ⎠ ( D − a)φ ( D ) xe ax 1 = e ax ∫ e ax ⋅ e − ax dx = ( F1 ) φ ( a) 1!φ ( a)

(

Ch03.indd 36

)

12/14/2011 11:33:02 AM

Linear Differential Equations with Constant Coefficients

( F2 ) f ′( a) ≠ 0

xe ax f ′( a)

or

3-37

Case (iii) f ( a) = 0, f ′( a) = 0, f ( r ) ( a) = 0, f ( r +1) ( a) ≠ 0, i.e., ( D − a) r is a factor of f ( D ) but ( D − a) r +1 is not a factor of f ( D ). Let f ( D ) = ( D − a) r φ ( D ) where ( D − a) is not a factor of φ ( D ). We have f ( r ) ( D ) = r !φ ( D ) + terms containing( D − a) as a factor ⇒ f ( r ) ( a) = r !φ ( a) ⎞ 1 1 1 ⎛ 1 e ax = e ax = e ax ⎟ ⎜ r r ( D − a) φ ( D ) φ ( a) ⎝ ( D − a) f ( D) ⎠ r ax 1 x e Thus, ( F3 ) e ax = f ( D) r !φ ( a) 1 x r e ax or ( F4 ) e ax = ( r ) f ( D) f ( a) D + 1 4x e . D2 − 5 Solution We have to evaluate Example 3.2.7 Evaluate

D + 1 4x e D2 − 5 D + 1 4x 4 + 1 4x 5 4x e = 2 e = e 2 D −5 4 −5 11 ∵ 4 is not a factor of ( D 2 − 5) . 1 e2 x . D −4 Solution Here f ( D ) = D 2 − 4 = ( D − 2)( D + 2) Example 3.2.8 Evaluate

∴

Ch03.indd 37

2

1 1 1 e2 x = e2 x = e2 x D −4 ( D − 2)( D + 2) ( D − 2)(2 + 2) 1⎛ 1 ⎞ 1 x 2x x 2x = ⎜ e2 x ⎟ = e = e . 4 ⎝ D − 2 ⎠ 4 1! 4 2

12/14/2011 11:33:03 AM

3-38

Differential Equations

Example 3.2.9 Evaluate Solution

1 e2 x . ( D − 2)2 ( D + 3)

Here, f ( D ) = ( D − 2)2 ( D + 3) = ( D − 2)2 φ ( D ) where φ ( D ) = D + 3 φ (2) = 2 + 3 = 5

1 x 2 e2 x x 2e2 x 2x e = = , by F3 ( D − 2)2 ( D + 3) 2! 5 10 Alternative method f ( D ) = ( D − 2)2 ( D + 3) = D 3 − D 2 − 8 D + 12 f ′ ( D ) = 3D 2 − 2 D − 8 f ′′( D ) = 3.2 D − 2.1 f ′′(2) = 3.2(2) − 2.1 = 10 ∴

1 1 e2 x = 3 e2 x 2 2 D − D − 8 D + 12 ( D − 2) ( D + 3) x 2e2 x x 2e2 x = = , by F4 . f ′′(2) 10

Example 3.2.10 Find the particular integral y p of ( D 2 − 3)3 y = e Solution

(

3x

.

)

f ( D ) = ( D 2 − 3) = D − 3 φ ( D ), 3

(

φ( D) = D + 3 yp =

( ∵ φ ( 3) = (

)

D− 3

3

1

)( ) 3

3

)

3

e

3x

D+ 3 1 x3 3+ 3 = ⋅ e 3 x! 2 3

( )

=

(

3x

1

)(

1

3

)

e

3+ 3 D− 3 x3 = e 3x , by F3 . 144 3

3x

,

Alternative method 3 f ( D ) = ( D 2 − 3) = D 6 − 9 D 4 + 27 D 2 − 27, f ′′′( D ) = 6.5.4 D 3 − 9.4.3.2 D

Ch03.indd 38

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Linear Differential Equations with Constant Coefficients

yp =

1 e f ( D)

=

3x

3

1 f ′′′

x e

= 6.5.4

( 3)

3

( )

e

3-39

3x

3

3x

− 9.4.3.2

( 3)

=

x 3 e 3x 144 3

Example 3.2.11 Solve y ′′ − y ′ − 2 y = 3e 2 x given that y(0) = 0, y ′(0) = −2.

[JNTU 2004] (3.40)

Solution To find the complementary function, we have to solve y ′′ − y ′ − 2 y = 0

(3.41)

The auxiliary equation is m2 − m − 2 = ( m − 2)( m + 1) = 0 ⇒ m = 2, −1 yc = c1e 2 x + c2 e − x

(3.42)

To find the particular integral, yp =

1 1 x Q( x ) = 2 3e 2 x = 3e 2 x = xe 2 x f ( D) D −D−2 (2 D − 1) D = 2

The general solution is y ( x ) = yc + y p = c1e 2 x + c2 e − x + xe 2 x

(3.43)

Differentiating Eq. (3.44) with respect to x y ′( x ) = 2c1e 2 x − c2 e − x + 1e 2 x + 2e2 x

(3.44)

y(0) = 0 = c1 ⋅ 1 + c2 ⋅ 1 + 0 ⇒ c1 + c2 = 0

(3.45)

y ′(0) = −2 = 2c1 ⋅ 1 + c2 ⋅ 1 + 1 + 0 ⇒ 2c1 − c2 + 1 = −2

(3.46)

Adding Eqs. (3.45) and (3.46) 3c1 + 1 = −2 ⇒ c1 = −1 (∵ c2 = 1) Substituting these values in Eq. (3.44), the required solution is obtained as y( x ) = − e 2 x + e − x + xe 2 x . (3.47)

Ch03.indd 39

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3-40

Differential Equations

Example 3.2.12 Solve ( D 3 + 2 D 2 + D ) y = e 2 x .

(3.48)

Solution To find the complementary function, we have to solve ( D3 + 2D 2 + D) y = 0

(3.49)

The auxiliary equation is m3 + 2m2 + m = m ( m + 1)2 = 0 ⇒ m = 0, −1, −1

(3.50)

yc = c1 + (c2 + c3 x ) e − x

(3.51)

To find the particular integral, yp =

1 1 1 e2 x 2x 2x Q ( x) = 3 e = e = f ( D) D + 2D 2 + D 23 + 2.22 + 2 18

The general solution is y ( x ) = yc + y p = c1 + (c2 + c2 x ) e − x +

e2 x . 18

(3.52)

Example 3.2.13 Solve ( D 3 − 1) y = (e x + 1)2 .

(3.53)

Solution To find the complementary function, we have to solve ( D 3 − 1) y = 0

(3.54)

The auxiliary equation is 1 i 3 m3 − 1 = ( m − 1)( m2 + m + 1) = 0 ⇒ m = 1, − ± 2 2 ⎛ 3 3 ⎞ −x 2 yc = c1e x + ⎜ c2 cos x + c3 sin x e 2 2 ⎟⎠ ⎝

(3.55)

To find the particular integral, 1 1 1 1 Q( x ) = 3 e2 x + 2 3 ex + 3 e 0. x f ( D) D −1 D −1 D −1 1 2x x (3.56) = e2 x + e −1 7 3

yp =

Ch03.indd 40

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Linear Differential Equations with Constant Coefficients

3-41

The general solution is ⎡ ⎛ 3 ⎞ ⎛ 3 ⎞ ⎤ −x 2 y( x ) = yc + y p = c1e x + ⎢c2 cos ⎜ x ⎟ + c3 sin ⎜ x⎟ ⎥ e ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎣⎢ 1 2x x (3.57) e − 1. + e2 x + 7 3

Example 3.2.14 Solve ( D 2 + 6 D + 9) y = e −3 x .

(3.58)

Solution To find the complementary function, we have to solve ( D 2 + 6 D + 9) y = 0 The auxiliary equation is m2 + 6m + 9 = ( m + 3)2 = 0 ⇒ m = −3, −3 yc = (c1 + c2 x ) e −3 x

(3.59)

To find the particular integral, 1 1 Q ( x) = 2 e −3 x f ( D) D + 6D + 9 x2 x2 ⎡⎣( D 2 + 6 D + 9) ″ ⎤⎦ D = −3 = 2 = e −3 x = e −3 x ; 2 2

yp =

The general solution is y( x ) = yc + y p = (c1 + c2 x ) e −3 x +

x 2 −3 x e 2

Example 3.2.15 Solve ( D 2 − k 2 ) y = cosh kx.

(3.60)

Solution To find the complementary function, we have to solve (D2 − k 2 ) y = 0

(3.61)

The auxiliary equation is m2 − k 2 = 0 ⇒ m = ± k yc = c1e kx + c2 e − kx

Ch03.indd 41

(3.62)

12/14/2011 11:33:04 AM

3-42

Differential Equations

To find the particular integral, 1 1 ⎛ 1 1 ⎞ (e kx + e − kx ) Q ( x) = − ⎜⎝ ⎟ f ( D) 2k D − k D + k ⎠ 2 1 ⎡ 1 1 1 1 ⎤ = e kx + e − kx − e kx − e − kx ⎥ ⎢ 4k ⎣ D − k D−k D−k D+k ⎦ 1 ⎡ kx 1 − kx 1 kx 1 x ⎤ sinh kx − 2 cosh kx = xe − e − e − xe − kx ⎥ = ⎢ 4k ⎣ 2k 2k 4k ⎦ 2k

yp =

The general solution is y ( x ) = yc + y p = c1e kx + c2 e − kx +

x 1 sinh kx − 2 cosh kx. 2k 4k

(3.63)

EXERCISE 3.5 Solve the following: 1. ( D 2 + 1) y = cosh x. 1 Ans: y = c1 cos x + c2 sin x + cosh x 2 2.

( D 2 − 1) y = sinh x. x Ans: y = c1 cosh x + c2 sinh x + cosh x 2

3. ( D + 1)3 y = e − x . Ans: y = ( c1 + c2 x + c3 x 2 ) e − x + 4.

( D 3 − 5D 2 + 7 D − 3) y = e2 x cosh x. Ans: y = (c1 + c2 x ) e x + c3e 3 x +

5.

x3 − x e 3!

x 3x (e − 8e x ) 8

( D 3 + 1) y = 3 + 5e x .

[JNTU 2002]

⎡ ⎛ 3 ⎞ ⎛ 3 ⎞⎤ x ⎟ + c3 sinh ⎜ x⎟ ⎥ Ans: y = c1 + ⎢c2 cos ⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎣⎢

Ch03.indd 42

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Linear Differential Equations with Constant Coefficients

3-43

6. (2 D + 1)2 y = 4e − x 2 . x2 − x 2 e 2

Ans: y = (c1 + c2 x ) e − x 2 + 7.

( D 2 − D − 6) y = e x cosh 2 x. Ans: y = c1e 3 x + c2 e −2 x +

8.

x 3x 1 − x e − e 10 8

( D 2 − 2 pD + p2 ) y = e x .

ex ⎧ px + + ( ) c c x e ⎪⎪ 1 2 ( p − 1)2 Ans: y = ⎨ 2 ⎪(c + c x ) e x + x e x ⎪⎩ 1 2 2! 4 x 9. ( D − 1) y = e .

( p ≠ 1) ( p = 1)

Ans: y = c1e x + c2 e − x + c3 cos x + c4 sin x +

x x e 4

10. (4 D 2 + 4 D − 3) y = e 2 x . Ans: y = c1e x 2 + c2 e −3 2 +

1 2x e 21

11. ( D 2 − 3D + 2) y = e 5 x . Ans: y = c1e x + c2 e 2 x +

1 5x e 12

12. ( D 2 + D + 1) y = e x. ⎛ x 3 x 3⎞ 1 x Ans: y = e − x 2 ⎜ c1 cos + c2 sin + e 2 2 ⎟⎠ 3 ⎝ 13. ( D 2 − 5 D + 6) y = 4e x + 5. Ans: y = c1e 2 x + c2 e 3 x + 2e x +

5 6

14. ( D 2 − 2 D + 1) y = (1 + e − x )2. 1 1 Ans: y = (c1 + c2 x )e x + 1 + e − x + e −2 x 2 9

Ch03.indd 43

12/14/2011 11:33:05 AM

3-44

Differential Equations

15. ( D 2 − 4) y = 3e 2 x − 4e −2 x. Ans: y = c1e −2 x + c2 e 2 x +

3 2x xe + xe −2 x 4

16. ( D 2 − 1) y = cosh x. x Ans: y = c1e x + c2 e − x + sinh x 2 (2) Q(x) = sin ax or cos ax where ‘a’ is a constant Case (i) f ( − a 2 ) ≠ 0 We have D (sin ax ) = a cos ax D 2 sin ax = D ( a cos ax ) = ( − a 2 ) sin ax

( D 2 )2 sin ax = ( − a2 )2 sin ax and generally,

( D 2 )r sin ax = ( − a2 )r sin ax for any integer ‘r’. ∴

f ( D 2 ) sin ax = f ( − a 2 ) sin ax

Operating on both sides by

1 , we have f ( D2 )

1 1 sin ax f ( D 2 ) sin ax = f ( − a 2 ) 2 f (D ) f ( D2 ) ⎛ 1 ⎞ sin ax ⎟ ⇒ sin ax = f ( − a 2 ) ⎜ 2 ⎝ f (D ) ⎠ ⇒

1 1 sin ax = sin ax 2 f (D ) f ( − a2 )

(3.64)

if f ( − a 2 ) ≠ 0.

Ch03.indd 44

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Linear Differential Equations with Constant Coefficients

3-45

Rule Replace D 2 by − a 2 if f ( − a 2 ) ≠ 0 The same rule applies if Q ( x ) = cos ax 1 1 cos ax = cos ax 2 f (D ) f ( − a2 )

(3.65)

if f ( − a 2 ) ≠ 0 This rule holds if sin ax and cos ax are replaced by sin ( ax + b) and cos( ax + b), where b is a constant. Case (ii) f ( − a 2 ) = 0, i.e.,

( D 2 + a2 ) f ( D 2 ) , ( D 2 + a2 )2 f ( D 2 )

The above rule fails if f ( − a 2 ) = 0. In this case, ( D 2 + a 2 ) f ( D 2 ) so that we can write f ( D 2 ) = ( D 2 + a 2 ) φ ( D 2 ) where φ ( − a 2 ) ≠ 0. Let us evaluate 1 1 ⎛ 1 1 ⎛ 1 1 ⎞ ⎞ e iax = eiax ⎟ = e iax ⎟ = ( xeiax ) ⎜⎝ ⎜ 2 ⎠ 2ia ⎝ D − ia ⎠ 2ia D +a D − ia D + ia 2

1 x (cos ax + i sin ax ) = (cos ax + i sin ax ) 2 D +a 2ia 2

Equating the real and imaginary parts, 1 x cos ax = sin ax 2 2a D +a 1 x sin ax = − cos ax 2 2 2a D +a 2

We have, if φ ( − a 2 ) ≠ 0,

Ch03.indd 45

12/14/2011 11:33:06 AM

3-46

Differential Equations

1 1 1 sin ax = 2 sin ax 2 2 f (D ) D + a φ ( D2 ) 1 ⎛ 1 ⎞ = sin ax⎟ 2 ⎜ 2 2 ⎠ φ ( −a ) ⎝ D + a 1 ⎛ x ⎞ = − cos ax⎟ 2 ⎜ ⎝ ⎠ φ ( − a ) 2a 1 1 1 cos ax = 2 cos ax 2 2 f (D ) D + a φ ( D2 ) 1 ⎛ 1 ⎞ = cos ax ⎟ 2 ⎜ 2 2 ⎠ φ ( −a ) ⎝ D + a 1 ⎛ x ⎞ = sin ax ⎟ . 2 ⎜ ⎝ ⎠ φ ( − a ) 2a

(3.66)

(3.67)

General case: If f ( D ) contains old powers of D, then we can arrange the even and odd powers of D so that f ( D ) = f1 ( D 2 ) + Df 2 ( D 2 ) Let Δ ( D 2 ) = ⎡⎣ f1 ( D 2 )⎤⎦ − D 2 ⎡⎣ f 2 ( D 2 )⎤⎦ 2

2

Then f1 ( D 2 ) − Df 2 ( D 2 ) 1 1 sin ax = sin ax = sin ax f ( D) f1 ( D 2 ) + Df 2 ( D 2 ) Δ ( D2 ) = ⎣⎡ f1 ( D 2 ) − Df 2 ( D 2 )⎦⎤

=

sin ax Δ ( − a2 )

if Δ ( − a 2 ) ≠ 0

f1 ( − a 2 ) sin ax − f 2 ( − a 2 ) a cos ax , if Δ ( − a 2 ) ≠ 0 Δ ( − a2 )

(3.68)

Similarly, f ( − a 2 ) cos ax + f 2 ( − a 2 ) a sin ax 1 cos ax = 1 . f ( D) Δ ( − a2 ) if Δ ( − a 2 ) ≠ 0

Ch03.indd 46

(3.69)

12/14/2011 11:33:06 AM

Linear Differential Equations with Constant Coefficients

Example 3.2.16 Evaluate Solution 1 D + D + D +1 3

2

1 sin 2 x. D3 + D2 + D + 1

1 sin 2 x ( D + 1) + D ( D 2 + 1) 1 D −1 sin 2 x = sin 2 x = −3( D + 1) −3( D 2 − 1) D (sin 2 x ) − 1 ⋅ sin 2 x 2cos 2 x − sin 2 x . = = 15 −3 ( −22 − 1)

sin 2 x =

Example 3.2.17 Evaluate Solution

3-47

2

1 sin 2 x. D −4 2

1 1 ⎛ 1 − cos 2 x ⎞ 1 1 1 1 sin 2 x = 2 e 0. x − ⋅ 2 cos 2 x ⎜⎝ ⎟⎠ = 2 D −4 D −4 2 2 D −4 2 D −4 1 1 1 1 1 1 = ⋅1 − ⋅ 2 cos 2 x = − + cos 2 x. 20−4 2 −2 − 4 8 16 2

Example 3.2.18 Solve ( D 2 − 3D + 2) y = sin 3 x.

(3.70)

Solution To find the complementary function, we have to solve ( D 2 − 3D + 2) y = 0

(3.71)

The auxiliary equation is m2 − 3m + 2 = 0 ⇒ ( m − 1)( m − 2) = 0 ⇒ m = 1,2 yc = c1e x + c2 e 2 x

(3.72)

To find the particular integral, 1 1 1 Q ( x) = 2 sin 3 x = 2 sin 3 x f ( D) D − 3D + 2 −3 − 3D + 2 1 3D − 7 sin 3 x. =− ⋅ 3D + 7 3D − 7

yp =

Ch03.indd 47

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3-48

Differential Equations

To get D 2 in the denominator 3D − 7 3D − 7 sin 3 x = − sin 3 x 2 2 9D − 7 9 ( −32 ) − 72 3( D sin 3 x ) − 7 ⋅ sin 3 x 9cos3 x − 7sin 3 x = = 130 130

=−

The general solution is y ( x ) = yc + y p = c1e x + c2 e x +

9cos3 x − 7sin 3 x . 130

Example 3.2.19 Solve ( D 2 − 1) y = sin x cos x. Solution To find the complementary function, we have to solve

( D 2 − 1) y = 0 The auxiliary equation is m2 − 1 = ( m − 1)( m + 1) = 0 ⇒ m = 1, −1 yc = c1e x + c2 e − x To find the particular integral, 1 1 1 1 yp = Q ( x) = 2 sin x cos x = sin 2 x f ( D) 2 D2 − 1 ( D − 1) 1 1 1 = sin 2 x = − sin 2 x 2 2 −2 − 1 10 The general solution is, y ( x ) = yc + y p = c1e x + c2 e − x −

1 sin 2 x. 10

Example 3.2.20 Solve ( D 2 − 4 D ) y = e x + sin 3 x cos 2 x. Solution To find the complementary function, we have to solve ( D 2 − 4 D) y = 0 The auxiliary equation is m2 − 4 m = 0 ⇒ m = 0, m = 4 yc = c1 + c2 e 4 x

Ch03.indd 48

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Linear Differential Equations with Constant Coefficients

3-49

To find the particular integral, yp =

1 1 1 ex + 2 ⋅ (sin 5 x + sin x ) D2 − 4D D − 4D 2

1 ∵ sin 3 x cos 2 x = [sin (3 x + 2 x ) + sin (3x − 2 x )] 2 1 1 1 ex = 2 ex = − ex D − 4D 1 − 4.1 3 1 1 4 D − 25 P .I 2 = 2 sin 5 x = 2 sin 5 x = − sin 5 x D − 4D −5 − 4 D 16 D 2 − 252 4.5cos5 x − 25sin 5 x 4 cos5 x − 5sin 5 x =− = 16 ( −52 ) − 625 105 1 1 P .I 3 = 2 sin x = 2 sin x D − 4D −1 − 4 D 4D − 1 1 =− sin x = (4 cos x − sin x ) 2 2 16 D − 1 17 1 x 4 cos5 x − 5sin 5 x 1 + (4 cos x − sin x ) yp = − e + 210 34 3 The general solution is P .I1 =

2

1 4 cos5 x − 5sin 5 x y( x ) = yc + y p = c1 + c2 e 4 x − e x + 3 210 1 + (4 cos x − sin x ). 34 Example 3.2.21 Solve ( D 3 + 2 D ) y = e 2 x + cos (3 x + 7). Solution To find the complementary function, we have to solve

( D3 + 2D ) y = 0 The auxiliary equation is m3 + 2 m = 0

⇒

(

m ( m2 + 2) = 0 ⇒ m = 0, m = ± 2i

yc = c1 + c2 cos 2 x + c3 sin 2 x

)

To find the particular integral, yp =

Ch03.indd 49

1 1 1 e2 x + 2 cos (3 x + 7) D + 2D D +2 D 3

12/14/2011 11:33:07 AM

3-50

Differential Equations

1 1 1 e2 x = 3 e2 x = e2 x D + 2D 2 + 2.2 12 1 1 1 1 1 P .I 2 . = sin (3x + 7) = . 2 (sin 3 x + 7) = − sin (3 x + 7) 2 3D +2 3 −3 + 7 6 e2 x 1 yp = − sin (3x + 7) 12 6 P . I1 . =

3

The general solution is

(

)

y ( x ) = yc + y p = c1 + c2 cos 2 x + c3 sin 2 x +

e2 x 1 − sin (3x + 7). 12 6

EXERCISE 3.6 Solve the following: 1. y ′′ + 4 y ′ + 47 = 4 cos x + 3sin x; y (0) = 1, y ′(0) = 0. Ans: y = (1 + x )e −2 x + sin x 2. ( D 3 − 1) y = e x + sin 3 x + 2.

[JNTU 2004]

⎡ ⎛ 3 ⎞ ⎛ 3 ⎞⎤ x ⎟ + c3 sin ⎜ x⎟ ⎥ Ans: y = c1e x + ⎢c2 cos ⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎣⎢ 27cos3 x − sin 3 x + 730 3.

( D 2 − 4) y = 2cos2 x.

1 Ans: y = c1e 2 x + c2 e −2 x − (2 + cos 2 x ) 8 2 4. ( D + 1) y = sin x sin 2 x.

[JNTU 2003]

cos x x Ans: y = c1 cos x + c2 sin x + sin x + 4 16 5.

( D 2 + 9) y = cos3x + sin 2 x. x sin 2 x Ans: y = c1 cos3 x + c2 sin 3 x + sin 3 x + 6 5

Ch03.indd 50

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Linear Differential Equations with Constant Coefficients

6.

3-51

( D 2 − 1) y = sin ( x + 5). 1 Ans: y = c1e x + c2 e − x − sin ( x + 5) 2

7.

( D 2 + 4) y = e x + sin 2 x + cos 2 x. 1 x Ans: y = (c1 cos 2 x + c2 sin 2 x ) + e x + ( − cos 2 x + sin 2 x ) 5 4

8.

( D 2 − 4 D + 3) y = sin 3x cos 2 x. Ans: y = c1e x + c2 e 3 x + +

1 (10 cos5 x − 11sin 5 x ) 884

1 (sin 2 x + 2cos x ) 20

(3) Q(x) = xm where m is a positive integer 1 1 x m , we expand To evaluate in increasing powers of D f ( D) f ( D) up to D m and apply on x m . Example 3.2.22 Evaluate Solution

1 ( x 2 + x + 1) . 1 + D + D2

1 1− D 2 ( x 2 + x + 1) = ( x + x + 1) 2 1+ D + D 1 − D3 = (1 − D )(1 − D 3 ) −1 ( x 2 + x + 1) = (1 − D ) (1 + D 3 + D 6 + )( x 2 + x + 1) = (1 − D + D 3 − )( x 2 + x + 1) = x 2 + x + 1 − (2 x + 1) = x 2 − x. Example 3.2.23 Evaluate ( D 2 + 3D ) −1 x. Solution

(D

Ch03.indd 51

2

+ 3D )

−1

−1

1 1 1 1 ⎛ D⎞ x= x= x= ⎜⎝1 + ⎟⎠ x D 2 3D + D 3D 1 + 3 3D 3 1 ⎛ D D2 ⎞ = − ⎟x ⎜1 − + ⎠ 3D ⎝ 3 9 11 1 1 1 1 1 = x− x+ Dx = x 2 − x + . 3D 9 27 6 9 27

12/14/2011 11:33:08 AM

3-52

Differential Equations

Example 3.2.24 Solve ( D 3 − 1) y = x 3 . Solution To find the complementary function, we have to solve ( D 3 − 1) y = 0 The auxiliary equation is m3 − 1 = 0

⇒ m = 1, 1 −

1 3 ± i 2 2

⎡ ⎛ 3 ⎞ ⎛ 3 ⎞ ⎤ −x 2 yc = c1e x + ⎢c2 cos ⎜ x ⎟ + c3 sin ⎜ x⎟ ⎥ e ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎣⎢ To find the particular integral, yp =

1 −1 x 3 = − (1 − D 3 ) x 3 = −1(1 + D 3 + D 6 + D −1 3

) x3 = − x3 − 6

The general solution is ⎡ ⎛ 3 ⎞ ⎛ 3 ⎞ ⎤ −x 2 y( x ) = yc + yc = c1e x + ⎢c2 cos ⎜ x ⎟ + c3 sin ⎜ x⎟ ⎥ × e − x 3 − 6. 2 2 ⎝ ⎠ ⎝ ⎠ ⎥⎦ ⎣⎢ Example 3.2.25 Solve ( D 3 − D 2 ) y = 1 + x 3 . Solution To find the complementary function, we have to solve ( D3 − D2 ) y = 0 The auxiliary equation is m3 − m2 = m2 ( m − 1) = 0 ⇒ m = 0,0,1 yc = c1 + c2 x + c3e x To find the particular integral, 1 1 (1 + x 3 ) = − 2 (1 − D ) −1 (1 + x 3 ) D3 − D2 D 1 = − 2 (1 + D + D 2 + D 3 + D 4 + D 5 + ) × (1 + x 3 ) D 1 ⎛ 1 ⎞ = − ⎜ 2 + + 1 + D + D 2 + D 3 + ⎟ × (1 + x 3 ) ⎝D ⎠ D 2 5 4 x x ⎛x ⎞ = −⎜ + x +1+ + + x 3 + 3 x 2 + 6 x + 6⎟ ⎝ 2 ⎠ 4.5 4

yp =

Ch03.indd 52

12/14/2011 11:33:09 AM

Linear Differential Equations with Constant Coefficients

3-53

The general solution is 7 ⎛ x5 x 4 ⎞ y( x ) = c1 + c2 x + c3e x − ⎜ + + x 3 + x 2 + 7 x + 7⎟ . ⎝ 20 4 ⎠ 2

Example 3.2.26 Solve (2 D 2 + D ) y = x 2 . Solution To find the complementary function, we have to solve (2 D 2 + D ) y = 0 The auxiliary equation is 2 m2 + m = 0 m (2m + 1) = 0

⇒

m = 0, −

1 y = c1 + c2 e x 2 2 c

To find the particular integral, 1 1 1 x 2 = (1 + 2 D ) −1 x 2 = (1 − 2 D + 4 D 2 − 8 D 3 + D (1 + 2 D ) D D x3 1 2 = ( x ) − 2 x 2 + 4 D( x 2 ) − 8D 2 ( x 2 ) = − 2 x 2 + 8 x − 16 D 3

yp =

) x2

The general solution is y( x ) = c1 + c2 e − x 2 +

x3 − 2 x 2 + 8 x − 16. 3

EXERCISE 3.7 Solve the following: 1. y ′′′ − 2 y ′′ − y ′ − 2 y = 1 − 4 x 3 .

[JNTU 2004(3)]

Ans: (5 + cx 2 ) x 3 y 5 = 2 2.

( D 2 + D + 1) y = x 3 .

[JNTU 2004]

⎡ ⎛ 3x ⎞ ⎛ 3x ⎞ ⎤ − x 2 + c2 sin ⎜ + x 3 − 3x 2 + 6 Ans: y = ⎢c1 cos ⎜ ⎟ ⎟⎥e 2 2 ⎝ ⎠ ⎝ ⎠ ⎥⎦ ⎢⎣ 3 2 3. ( D − 3D − 2) y = x . Ans: y = c1e 2 x + (c2 + c3 x )e − x − 14 (2 x 2 + 6 x + 9)

Ch03.indd 53

12/14/2011 11:33:09 AM

3-54

Differential Equations

4. ( D 3 − D 2 − D + 1) y = 1 + x 2 .

[JNTU 1997S]

Ans: y = c1e − x + (c2 + c3 x ) e x + x 2 + 2 x + 5 5. ( D 3 + 2 D 2 + D ) y = e 2 x + x 2 + x + sin 2 x. Ans: y = c1 + (c2 + c3 x )e − x + +

[JNTU 2003S(1)]

e2 x x + (2 x 2 − 9 x + 24) 18 6

1 (3cos 2 x − 4sin 2 x ) 50

6. ( D 2 + 5 D + 4) y = x 2 . Ans: y = c1e − x + c2 e −4 x +

[JNTU 2003S(2)] 1 (8 x 2 − 20 x + 21) 32

(4) Q(x) = eaxV(x) where ‘a’ is a constant and V(x) is a function of x By the theorem on exponential shift, we have f ( D ) (e axV1 ) = e a x f ( D + a)V1

(3.73)

where V1 is a function of x. Applying

1 on both sides, we have f ( D) 1 e axV1 = [e ax f ( D + a)V1 ] f ( D) ⇔

1 [e ax f ( D + a)V1 ] = e axV1 f ( D)

(3.74) (3.75)

Let f ( D + a)V1 = V ⇔ V1 =

1 V f ( D + a)

(3.76)

From Eqs. (3.75) and (3.76), we get 1 1 V (e axV ) = e ax f ( D) f ( D + a)

The effect of shifting the exponential factor e ax from the right side of the operator to the left side is to replace D by ( D + a) in f.

Ch03.indd 54

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Linear Differential Equations with Constant Coefficients

Example 3.2.27 Evaluate Solution

3-55

1 e x x. ( D − 1)3

1 1 1 x4 x4e x x x x x e x = e x = e x = e ⋅ = . ( D − 1)3 ( D + 1 − 1)3 D3 2.3.4 24 Example 3.2.28 Evaluate Solution

1 xe 2 x . D − 4D + 4 2

1 1 1 e2 x ⋅ x = e2 x x = e2 x x 2 D − 4D + 4 ( D − 2) ( D + 2 − 2)2 x3 x 3e x 1 = e2 x 2 x = e2 x = . D 2⋅3 6 2

Example 3.2.29 Evaluate Solution

1 e 5 x sin x. ( D − 5)3

1 1 1 e 5 x sin x = e 5 x sin x = e5 x 3 sin x = − e5 x cos x. 3 3 D ( D − 5) ( D + 5 − 5) Example 3.2.30 Solve ( D 4 − 1) y = e x cos x. Solution To find the complementary function, we have to solve ( D 4 − 1) y = 0 The auxiliary equation is m4 − 1 = ( m − 1)( m + 1) ( m2 + 1) = 0 ⇒ m = ±1, ±i yc = c1e x + c2 e − x + c3 cos x + c4 sin x

To find the particular integral, 1 1 e x cos x = e x cos x D −1 ( D + 1) 4 − 1 1 = ex 4 cos x 3 D + 4 D + 6D 2 + 4 D e x cos x 1 = ex x = − cos 5 ( −12 )2 + 4 ( −12 ) D + 6 ( −12 ) + 4 D

yp =

Ch03.indd 55

4

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3-56

Differential Equations

The general solution is y( x ) = yc + y p = c1e x + c2 e − x + c3 cos x + c4 sin x −

e x cos x . 5

Example 3.2.31 Solve ( D 2 + 4 D + 4) y = e −2 x cos x. Solution To find the complementary function, we have to solve ( D 2 + 4 D + 4) y = 0 The auxiliary equation is m2 + 4 m + 4 = ( m + 2)2 = 0 ⇒ m = −2, −2 yc = (c1 + c2 x ) e −2 x To find the particular integral, 1 e −2 x −2 x = e cos x cos x ( D + 2)2 ( D − 2 + 2)2 1 = e −2 x 2 cos x = −e −2 x cos x D

yp =

The general solution is y( x ) = y ( x ) = yc + y p = (c1 + c2 x ) e −2 x − e −2 x cos x. EXERCISE 3.8 Solve the following: 1. ( D 2 − 7 D + 6) y = e 2 x (1 + x ). Ans: y = c1e − x + c2 e −6 x −

[JNTU 2003]

2x

e (4 x + 1) 16

2. ( D 2 − 2 D + 1) y = xe x . Ans: y = (c1 + c2 x ) e x − e x ( x sin x + 2cos x ) 3. ( D 3 − 3D 2 + 3D − 1) y = x 2 e x . Ans: y = (c1 + c2 x + c3 x 2 )e x +

Ch03.indd 56

[JNTU 2002S] x 5e x 60

12/14/2011 11:33:10 AM

Linear Differential Equations with Constant Coefficients

4. ( D 2 + 4 D + 3) y = e − x sin x + x. Ans: y = c1e − x + c2 e −3 x −

3-57

[JNTU 1996, 1999]

−x

e 1⎛ 4⎞ (2cos x + sin x ) + ⎜ x − ⎟ ⎝ 5 3 3⎠

5. ( D 2 − 4) y = x sinh x.

[JNTU 2000]

x 2 Ans: y = c1e 2 x + c2 e −2 x − sinh x − cosh x 3 9 6. ( D 3 − 7 D 2 + 14 D − 8) y = e x cos 2 x. Ans: y = c1e x + c2 e 2 x + c3e 4 x +

[JNTU 2003S(3)]

2e x (8cos 2 x − sin 2 x ) 229

7. ( D 2 − 4 D + 1) y = e 2 x cos3 x. Ans: y = e 2 x (c1 cos3 x + c2 sin 3 x ) +

xe 2 x sin 3x 6

(5) Q(x) = xV(x) where V(x) is a function of x Theorem 3.2.32 Prove that D n ( xV1 ) = xD nV1 + nD n −1V1

(3.77)

where V1 is a function of x. Proof For n = 1 D1 ( xV1 ) = x ( DV1 ) + V1 (true) Assume that Eq. (3.77) is true for n = m D m ( xV1 ) = xD mV1 + mD m −1V1

(3.78)

Applying D on both sides, D m+1 ( xV1 ) = D ( xD mV1 ) + mD mV1 = xD m+1V1 + ( m + 1) D mV1

(3.79)

If Eq. (3.77) is true for n = m, then it is true for n = m + 1. Hence by the principle of mathematical induction, Eq. (3.77) is true for all n ∈N . Let f ( D ) = ∑ Cr D r

(

)

f ( D )( xV1 ) = ∑ Cr D r ( xV1 ) = x = xf ( D )V1 + f ′( D )V1

Ch03.indd 57

(∑ C D ) V + ∑ rC D r

r

1

r

r −1

V1

12/14/2011 11:33:10 AM

3-58

Differential Equations

Take V1 =

1 V in the above result. f ( D)

⎛ ⎞ 1 1 We have f ( D ) ⎜ x V ⎟ = xV + f ′( D ) V f ( D) ⎝ f ( D) ⎠ 1 1 1 1 ( xV ) + V= f ′( D ) V f ( D) f ( D) f ( D) f ( D) 1 1 1 1 ( xV ) = x V− f ′( D ) V ⇒ f ( D) f ( D) f ( D) f ( D) ⎛ ⎞ 1 1 =⎜x− f ′ ( D )⎟ V. f ( D) ⎝ ⎠ f ( D) ⇒ x

Example 3.2.33 Solve ( D + 1)2 y = x cos x. Solution To find the complementary function, we have to solve ( D + 1)2 y = 0. The auxiliary equation is ( m + 1)2 = 0 ⇒ m = −1, −1 yc = (c1 + c2 x ) e − x To find the particular integral, 1 yp = x cos x ( D + 1)2 Here f ( D ) = ( D + 1)2 f ′( D ) = 2( D + 1) ⎡ ⎤ 1 1 ( x cos x ) = ⎢x − f ′( D ) ⎥ f ( D) ⎣ ⎦ f ( D) ⎡ 2( D + 1) ⎤ 1 cos x = ⎢x − 2 ⎥ 2 ( D + 1) ⎦ −1 + 2 D + 1 ⎣ 2 ⎞ sin x x D −1 ⎛ sin x =⎜x− = sin x − 2 ⎟ ⎝ 2 D + 1⎠ 2 D −1 cos x − sin x x = sin x + 2 2 The general solution is x 1 y ( x ) = yc + y p = (c1 + c2 x ) e − x + sin x + (cos x − sin x ). 2 2

Ch03.indd 58

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Linear Differential Equations with Constant Coefficients

3-59

Example 3.2.34 Solve ( D 2 + 9) y = x sin 2 x. Solution To find the complementary function, we have to solve ( D 2 + 9) y = 0. The auxiliary equation is m2 + 9 = 0 ⇒ m = ±3i yc = c1 cos3 x + c2 sin 3 x To find the particular integral, yp =

1 x sin x D +9 2

Here f ( D ) = D 2 + 9 f ′( D ) = 2 D ⎡ ⎤ 1 1 2D ⎞ 1 ⎛ sin 2 x = ⎜ x − 2 sin 2 x f ′( D ) ⎥ = ⎢x − ⎟ ⎝ ( ) ( ) f D f D D + 9⎠ D2 + 9 ⎣ ⎦ 2D ⎞ 1 2.2cos 2 x x ⎛ sin 2 x = sin 2 x − =⎜x− 2 ⎟ ⎝ 5 5( −4 + 9) D + 9 ⎠ −4 + 9 x sin 2 x 4 cos 2 x x sin 2 x 4 = − = − cos 2 x 5 5( −4 + 9) 5 25 The general solution is y( x ) = yc + y p = c1 cos3 x + c2 sin 3 x +

x sin 2 x 4 − cos 2 x. 5 25

Example 3.2.35 Solve ( D − 2)2 y = x 2 sin x + e 2 x + 3. [JNTU 2001S] Solution yc = (c1 + c2 x )e 2 x 1 1 x 2 sin x = x ( x sin x ) 2 ( D − 2) ( D − 2)2 ⎡ ⎤ ⎡ D−2 ⎤ 1 ( x sin x ) ⎥ − 2 ⎢ (sin x ) ⎥ = xI1 − 2 I 2 = x⋅⎢ 2 4 ⎣ ( D − 2) ⎦ ⎣ ( D − 2) ⎦ 1 1 ( D − 2) ( x sin x ) = x ⋅ (sin x ) − 2 ⋅ (sin x ) I1 = 2 2 ( D − 2) ( D − 2) ( D − 2) 4 1 1 = x⋅ (sin x ) − 2 ⋅ (sin x ) 3 − 4D ( D − 2)3

P .I1 =

Ch03.indd 59

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3-60

Differential Equations

3 + 4D 11D + 2 (sin x ) − 2 ⋅ 2 2 (sin x ) 32 − 4 2 D 2 11 D − 22 ( D − 2)3 = D 3 − 3 ⋅ 2 ⋅ D 2 + 3 ⋅ 22 D − 23 = 11D − 2, Putting D 2 = −1 4 ( D − 2) = D 4 − 4 ⋅ 2 D 3 + 6 ⋅ 22 D 2 − 4 ⋅ 23 = −7 − 24 D , Putting D 2 = −1 2 x = (3sin x + 4 cos x ) + (2sin x + 11cos x ) 25 125 1 1 1 (sin x ) ( x sin x ) = x (sin x ) − 3 ⋅ I2 = 3 3 ( D − 2) 4 ( D − 2) ( D − 2) 3 x = (11D + 2)(sin x ) − (7 − 24 D )(sin x ) −125 625 3 x =− (2sin x + 11cos x ) + (7sin x − 24 cos x ) 125 625 2x x2 (2sin x + 11cos x ) P.I1 = xI1 − 2 I 2 = (3sin x + 4 cos x ) + 25 125 2x 6 + (2sin x + 11cos x ) − (7sin x − 24 cos x ) 125 625 1 = [(75 x 2 + 40 x − 42)sin x + (100 x 2 + 220 x + 144) cos x ] 625 1 x2 2x 2x P .I 2 . e = e ⋅ ( D − 2)2 2 1 3 P .I 3 . 3= ( D − 2)2 4 y p = P .I1 + P .I 2 + P .I 3 = x⋅

The general solution is y = yc + y p = (c1 + c2 x )e 2 x x2 3 1 + e 2 x + + [(75 x 2 + 40 x − 42)sin x 2 4 65 +(100 x 2 + 220 x + 144) cos x ] Alternatively, P .I1 can also be obtained as follows: P . I1 =

Ch03.indd 60

1 1 x 2 sin x = x 2 ( Imeix ) ( D − 2)2 ( D − 2)2

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Linear Differential Equations with Constant Coefficients

3-61

1 (by exponential shift) ( D + i − 2)2 1 = Ime ix x2 [(2 − i ) − D ]2 −2 D ⎞ 1 ⎛ 2 = Imeix − 1 ⎜ ⎟ x (2 − i )2 ⎝ 2 − i ⎠ ⎤ 2 ⎡ 1 1 = Imeix 1+ 2D + 3D 2 + ⎥ x 2 2 ⎢ 2 (2 − i ) ⎣ (2 − i ) (2 − i ) ⎦ (by Binomial theorem) −7 + 24i ⎤ ⎡ 3 + 4i 2 2 + 11i = Imeix ⎢ 4x + .6⎥ x + 125 625 ⎣ 25 ⎦ = Ime ix

1 2+i 1 1 3 + 4i = = = 2 2−i 5 (2 − i ) 3 − 4i 25 1 2 + 11i 1 −7 + 24i = = (2 − i )3 125 (2 − i ) 4 625 = Im (cos x + i sin x )[(75 + 100i ) x 2 + (40 + 220i ) + ( −42 + 144i )] 1 [(75 x 2 + 40 x − 42)sin x + (100 x 2 + 220 x + 144) cos x ]. = 625 EXERCISE 3.9 Solve the following: 1. ( D + 1)2 y = x cos x.

[JNTU 2003]

1 Ans: = (c1 + c2 x ) e − x + ( x sin x + cos x − sin x ) 2 2. ( D 2 + 4) y = x sin x.

[JNTU 2003]

1⎛ 2 ⎞ Ans: y = c1 cos 2 x + c2 sin 2 x + ⎜ x sin x − cos x ⎟ ⎠ 3⎝ 3

3. ( D 2 − 1) y = x sin 3 x + cos x. Ans: y = c1e x + c2 e − x −

Ch03.indd 61

[JNTU 1997]

1 ⎛3 ⎞ ⎜⎝ cos3 x + x sin 3 x ⎟⎠ 10 5

12/14/2011 11:33:12 AM

3-62

4.

Differential Equations

( D 2 + 1) y = x 2 sin x.

1 1 Ans: y = − cos x − ( 2 x 3 cos x − 3 x cos x − 3 x 2 sin x ) 2 12 [JNTU 2002] 5. ( D − 1)2 y = xe x sin x. Ans: y = (c1 + c2 x ) e x − e x (2cos x + x sin x ) 6.

( D 2 − 1) y = x sin x + x 2e x . 1 x Ans: y = c1e x + c2 e − x − ( x sin x + cos x ) + e x ( 2 x 2 − 3 x + 3) 2 12

7.

( D 2 + 1) y = x 2 sin x.

1 (2 x 3 cos x − 3x cos x − 3x 2 sin x ) 12 8. ( D 2 − 1) y = x sin x + e x (1 + x 2 ) . [JNTU 199S] Ans: y = c1 cos x + c2 sin x −

1 Ans: y = c1e x + c2 e − x − (cos x + x sin x ) 2 xe x e x ⎛ x 3 x 2 x ⎞ + + ⎜ − + ⎟ 2 2⎝ 3 2 2⎠ 9.

( D 2 + 3D + 2) y = xe x sin x.

Ans: y = (c1e − x + c2 e −2 x ) +

ex [5 x(sin x − cos x ) 5 −2sin x + 5cos x ]

10.

( D 2 + 1) y = e − x + x 3 + e x sin x.

[JNTU 1998]

1 −e x Ans: y = c1 cos x + c2 sin x + e − x (2cos x − sin x ) + x 3 − 6 x 2 5 11.

( D 2 + 2 D + 1) y = −

e− x . x2

Ans: y = (c1 + c2 x ) e − x − e − x log x 12. y ′′ + 4 y ′ + 20 y = 23sin t − 15cos; y(0) = 0, y ′(0) = 1. [JNTU 2004(4)] Ans: y ± =

Ch03.indd 62

e −2t (2cos 4t + sin 4t ) + sin t − cos t 2

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Linear Differential Equations with Constant Coefficients

13.

( D 3 − 1) y = e x + sin 3x + 2.

3-63

[JNTU 2004(1)]

⎛ ⎛ 3 ⎞ ⎛ 3 ⎞⎞ x⎟ + c3 sin ⎜ x⎟ ⎟ Ans: y = c1e x + e − x 2 ⎜ c2 cos ⎜ ⎝ 2 ⎠ ⎝ 2 ⎠⎠ ⎝ x 1 (27cos3 x − sin 3 x ) − 2 + ex + 3 730 14. ( D 2 + 1) x = t cos 2t x = x ′ = 0 at t = 0. [JNTU 2002] 5 t 4 Ans: x(t ) = − sin t − cos 2t + sin 2t 9 3 9 15.

( D 2 − 4 D + 4) y = 8 x 2 e2 x sin 2 x.

[JNTU 2001S]

Ans: y = (c1 + c2 x ) e 2 x − 2e 2 x ⋅ [2 x cos 2 x + ( x 2 − 3 2)sin 2 x ]

3.2.2 Linear Equations with Variable Coefficients— Euler–Cauchy Equations (Equidimensional Equations) An equation of the form f ( xD ) y ≡ ( a0 x n D n + a1 x n −1 D n −1 + or

+ an −1 xD + an ) y = Q ( x )

(3.80)

⎛ n ⎞ n−r n−r ⎜⎝ ∑ ar x D ⎟⎠ y = Q ( x ) r=0

where ai are all constants and Q ( x ) is a function of x is called an Euler–Cauchy equation. Some authors call it a homogeneous equation. Since the phrase is used for an equation with Q ( x ) ≡ 0, it leads to confusion if we use it for Eq. (3.80). Equation (3.80) is rightly called an equidimensional equation since the dimension of each term with respect to the independent variable is same. Change the independent variable to z by putting x = e z or z = log x so that ⎫ ⎪ dy dy dz 1 dy = = ⇒ xDy = θ y ⎪ dx dz dx x dz ⎪ d 2 y 1 d 2 y dz 1 dy 1 ⎛ d 2 y dy ⎞ ⎪ = − = − ⎜ ⎟⎬ dx 2 x dz 2 dx x 2 dz x 2 ⎝ dz 2 dz ⎠ ⎪ ⇒ x 2 D 2 y = θ (θ − 1) y ⎪ ⎪ By induction ⎪ x r D r y = θ (θ − 1) (θ − r + 1) y ⎭

Ch03.indd 63

(3.81)

12/14/2011 11:33:13 AM

3-64

Differential Equations

d d and θ = dx dz Substituting these into Eq. (3.80) the equation reduces to a linear differential equation with constant coefficients, which can be solved by the methods discussed above. where D =

dy + y = x. dx Solution Put x = e z , xDy = θ y d d where D = , θ = dx dz To equation reduces to Example 3.2.36 Solve x

(θ + 1) y = e z

(3.82) (3.83) (3.84)

(3.85)

To find the complementary function, we solve (θ + 1) y = 0

(3.86)

The auxiliary equation is m +1= 0

⇒

m = −1 c yc = ce − z = x

(3.87)

To find the particular integral, yp −

1 z 1 z x e = e = θ +1 1+1 2

(3.88)

c x + . x 2

(3.89)

The general solution is y=

Example 3.2.37 Solve 2 x 2

d2 y dy + 3 x − y = x. 2 dx dx

Solution Put x = e z or z = log x so that x

dy d2 y = θ y, x 2 2 = θ (θ − 1) y dx dx

where θ =

Ch03.indd 64

(3.90) (3.91) (3.92)

d dz

12/14/2011 11:33:13 AM

Linear Differential Equations with Constant Coefficients

3-65

The given equation becomes [2θ (θ − 1) + 3θ − 1] y = e z or (2θ 2 + θ − 1) y = e z

(3.93)

which is a linear equation with constant coefficients. To find the complementary function, we have to solve (2θ 2 + θ − 1) y = 0

(3.94)

The auxiliary equation is 2 m2 + m − 1 = 2 m2 + 2 m − m − 1 = 0 ( m + 1)(2m − 1) 1 m = −1 = 0 or yc = c1e − z + c2 e z 2 2 1 = c1 + c2 x x

(3.95) (3.96) (3.97)

To find the particular integral, 1 1 ez x = yp = 2 ez = ez = 2 2θ + θ − 1 2 ×1 +1−1 2 2 The general solution is y = c1 Example 3.2.38 Solve x 2

1 x + c2 x + . x 2

d2 y dy + 3 x + y = log x. 2 dx dx

Solution Put x = e z or z = log x

(3.99)

d2 y = θ (θ − 1) y dx 2 d where θ= . dz The given equation becomes so that

x

dy = θ y, dx

(3.98)

x2

[θ (θ − 1) + 3θ + 1]y = z or (θ + 1)2 y = z which is a linear equation with constant coefficients. To find the complementary function, we have to solve (θ + 1)2 y = 0

(3.100)

(3.101)

(3.102)

The auxiliary equation is ( m + 1)2 = 0

Ch03.indd 65

⇒

m = −1, −1

12/14/2011 11:33:14 AM

3-66

Differential Equations

yc = (c 1 + c2 z )e − z = (c1 + c2 log x )

1 x

(3.103)

To find the particular integral, 1 z = (1 + θ ) −2 z (θ + 1)2 = (1 − 2θ + 3θ 2 − ) z = z − 2 = log x − 2

yp =

(3.104)

The general solution is 1 y = (c1 + c2 log x ) + log x − 2. x Example 3.2.39 Solve x 2

d2 y dy − x + 2 y = x log x. 2 dx dx

(3.105)

Solution Put x = e z or z = log x ⎫ ⎪ 2 dy d y d 2 so that x = θ y, x = θ (θ − 1) y θ = ⎬⎪ 2 dx dx dz ⎭

(3.106)

The given equation becomes [θ (θ − 1) − θ + 2] y = ze z or (θ 2 − 2θ + 2) y = ze z

(3.107)

which is a linear equation with constant coefficients. To find the complementary function, we have to solve (θ 2 − 2θ + 2) y = 0

(3.108)

The auxiliary equation is m2 − 2m + 2 = ( m − 1)2 + 1 = 0 ⇒ m − 1 = ± i ⇒ m = 1 ± i yc = (c1 cos z + c2 sin z )e z = [c1 cos(log x ) + c2 sin(log x )]x

(3.109) (3.110)

To find the particular integral, 1 1 ze z = e z z (exponential shift) 2 (θ − 1) + 1 (θ + 1 − 1) 2 + 1 = e z (1 + θ )2 z = e z (1 − θ 2 + ) z = ze z = x log x

yp =

Ch03.indd 66

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Linear Differential Equations with Constant Coefficients

3-67

The general solution is y = [c1 cos(log x ) + c2 sin(log x )]x + x log x. Example 3.2.40 Solve ( x 2 D 2 + xD + 1) y = log x sin (log x ). (3.111) Solution Put x = e z or z = log x

(3.112)

so that xD = θ , x D = θ (θ − 1) The given equation becomes 2

2

[θ (θ − 1) + θ + 1] y = z sin z (θ 2 + 1) y = z sin z

(3.113)

which is a linear equation with constant coefficients. To find the complementary function, we have to solve (θ 2 + 1) y = 0 The auxiliary equation is m2 + 1 = 0

⇒

m = ±i

yc = c1 cos z + c2 sin z = c1 cos(log x ) + c2 sin(log x )

(3.114) (3.115)

To find the particular integral, 1 1 1 z sin z = Im 2 ze iz = Ime iz 2 z θ +1 θ +1 (θ + 1)2 + 1 (exponential shift) − 1 θ⎞ e iz ⎛ 1 = Ime iz 2 z = Im ⎜⎝1 + ⎟⎠ z θ + 2iθ 2iθ 2i 2 iz θ θ e ⎛ −i ⎞ ⎛ 1 1 θ ⎞ ⎞ iz ⎛ = Im ⎟z ⎜⎝1 − + 2 − ⎟⎠ z = Ime ⎜⎝ ⎟⎠ ⎜⎝ − − 2i θ 2i 4i 2 θ 2i 4 ⎠ ⎛ i ⎞ ⎛ z2 z 1 ⎞ = Ime iz ⎜ − ⎟ ⎜ − − ⎟ ⎝ 2 ⎠ ⎝ 2 2i 4 ⎠ ⎡ ⎛ 1 z2 ⎞ z ⎤ = Im(cos z + i sin z ) ⎢i ⎜ − ⎟ + ⎥ ⎣ ⎝ 8 4 ⎠ 4⎦ ⎛ ⎞

yp =

Ch03.indd 67

2

12/14/2011 11:33:15 AM

3-68

Differential Equations

z ⎛ 1 z2 ⎞ = ⎜ − ⎟ cos z + sin z ⎝8 4 ⎠ 4 1 ⎛1 1 ⎞ = ⎜ − (log x )2 ⎟ cos(log x ) + (log x )sin(log x ) ⎝8 4 ⎠ 4 The general solution is ⎛ 1 ⎞ y = c1 cos(log x ) + c2 sin(log x ) + ⎜ − (log x )2 ⎟ cos(log x ) ⎝ 4 ⎠ 1 + (log x )sin(log x ). 4 ⎛1 ⎞ ⎜⎝ cos(log x )is merged in the complementary function.⎟⎠ 8 d3 y d2 y 1⎞ ⎛ 2 + 2 x + 2 y = 10 ⎜ x + ⎟ . (3.116) 3 2 ⎝ dx dx x⎠ or z = log x

Example 3.2.41 Solve x 3 Solution Put x = e z so that x

dy d2 y = θ y, x 2 2 = θ (θ − 1) y, dx dx

d3 y = θ (θ − 1)(θ − 2) y dx 3 The given equation becomes x3

[θ (θ − 1)(θ − 2) + 2θ (θ − 1) + 2] y = 10 ( e z + e − z ) (θ 3 − θ 2 + 2) y = 10 ( e z + e − z )

(3.117) (3.118)

(3.119)

which is a linear equation with constant coefficients. To find the complementary function, we have to solve

(θ 3 − θ 2 + 2) y = 0 The auxiliary equation is m3 − m2 + 2 = 0 ⇒ m = −1 is a root

By synthetic division m2 − 2m + 2 = ( m − 1)2 + 1 = 0 ⇒ m = 1 ± i

Ch03.indd 68

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Linear Differential Equations with Constant Coefficients

yc = c1e − z + (c2 cos z + c3 sin z )e z 1 = c1 + [c2 cos(log x ) + c3 sin(log x )]x x To find the particular integral, y p = 10

1

θ −θ + 2 3

2

e z + 10

1

θ −θ 2 + 2 3

1 1 e z + 10 z 2 e− z 2 1 −1 + 2 3θ − 2θ 2 = 5e z + 2 ze − z = 5 x + log x x The general solution is = 10

y = c1

e− z

3-69

(3.120)

(3.121)

3

(3.122)

1 2 + [c2 cos(log x ) + c3 sin (log x )]x + 5 x + log x. (3.123) x x

EXERCISE 3.10 Solve the following: d2 y 1 1. x 2 2 − 2 y = x 2 + . dx x Ans: y = c1 x 2 +

c2 x

+ 13 ( x 2 − 1x ) log x

d3 y d2 y dy − 4 x 2 + 6 = 4. 3 dx dx dx 2 Ans: y = c1 + c2 x 3 + c3 x 4 + x 3

2. x 2

[Hint: Multiply by x]

d2 y dy + 2 x − 20 y = ( x + 1)2 . dx 2 dx x2 Ans: y = c1 x 4 + c2 x15 − 14 − 9x − 201

3. x 2

d2 y dy − 4 x + 6 y = x2 . 2 dx dx 2 Ans: y = c1 x + c2 x 3 − x 2 log x

4. x 2

Ch03.indd 69

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3-70

Differential Equations

d2 y dy 1 − x + y = [log x sin (log x ) + 1]. dx 2 dx x 1 6 + cos(log x ) Ans: y = x(c1 + c2 log x ) + 100 x 125 x 4 1 + sin(log x ) + log x[4 cos(log x ) + 3sin(log x )] 125 x 25 x d2 y dy 1 6. x 2 2 + 3 x + y = . dx dx (1 − x )2 1 1 x Ans: y = (c1 + c2 log x ) + log x x (1 − x ) 2 d y dy 7. x 2 2 + x − 3 y = x 2 . dx dx Ans: y = c1 x 3 + c2 x − 3 + x 2 5. x 2

d2 y dy + 8 x + 12 y = x 4 . 2 dx dx x4 Ans: y = c1 x −3 + c2 x −4 + 16 2 d y dy x 2 2 − 3 x − 5 y = sin(log x ). dx dx c1 1 Ans: y = + c2 x 5 + [2cos(log x ) − 3sin(log x )] x 26 2 d y dy x 2 2 + 7 x + 13 y = log x. dx dx 1 1⎛ 1⎞ Ans: y = 3 [c1 cos(2log x ) + c2 sin(2log x )] + ⎜ log x − ⎟ x 13 ⎝ 13 ⎠ 2 d y dy x 2 2 + x + y = x + log x. dx dx x Ans: y = c1 cos(log x ) + c2 sin(log x ) + + log x 2 2 d y dy x 2 2 − 4 x + 6 y = x. dx dx Ans: y = c1 x 2 + c2 x 3 + 2x d2 y dy x 2 2 − x + 2 y = x log x. [JNTU 1993] dx dx

8. x 2

9.

10.

11.

12.

13.

Ans: y = x(c1 cos(log x ) + c2 sin(log x )) + x log x

Ch03.indd 70

12/14/2011 11:33:16 AM

Linear Differential Equations with Constant Coefficients

14. x 2

d2 y dy − 2 x − 4 y = x 2 + 2log x. dx 2 dx

Ans: y = c1 x 4 + 15. x 2

3-71

c2 x

−

x2 6

− 12 ( log x +

3 4

)

d2 y dy − x + y = log x. 2 dx dx

[JNTU 1995]

Ans: y = (c1 + c2 log x ) x + log x + 4 16. (1 + x )2

d2 y dy + (1 + x ) + y = sin[2log(1 + x )]. 2 dx dx

[JNTU 1996]

Ans: y = c1 cos log(1 + x ) + c2 sin log(1 + x ) − 13 sin[2log(1 + x )] d3 y d2 y dy 1 + 2 x3 2 − x2 + xy = . 3 dx dx dx x c 1 Ans: y = 1 + (c2 + c3 log x ) x − 2 x 8x

17. x 4

[JNTU 2004]

3.2.3 Legendre’s Linear Equation An equation of the form d n−r ⎞ dn y n ( ) = + y a a bx ⎟ r 0 dx n − r ⎠ dx n r=0 n −1 d y + a1 ( a + bx ) n −1 n −1 + dx dy + an −1 ( a + bx ) + an y = Q ( x ) dx n

⎛

∑ ⎜⎝ a (a + bx)

n−r

(3.124)

where ar ( r = 0,1, , n), a, b are constants and Q ( x ) is a function of x is called a Legendre’s linear equation. Put a + bx = e z or z = log( a + bx ) so that dy dy dz b dy d2 y = = ⇒ ( a + bx ) = bθ y 2 dx dz dx a + bx dx dx 2 b2 d y ⎛ d 2 y dy ⎞ = − ⎟ ⇒ ( a + bx )2 2 = b 2θ (θ − 1) y ⎜ ( a + bx )2 ⎝ dz 2 dz ⎠ dx

Ch03.indd 71

12/14/2011 11:33:16 AM

3-72

Differential Equations

By induction, ( a + bx ) r

dr y = b rθ (θ − 1) dx r

(θ − r + 1) y

Equation (3.69) now reduces to a linear equation with constant coefficients which can be solved by the methods discussed above. Example 3.2.42 Solve (3 x + 2)2

d2 y dy + 3(3 x + 2) − 36 y = 3 x 2 + 4 x. 2 dx dx

(3.125)

Solution This is Legendre’s linear equation. Put 3 x + 2 = e z or z = log(3 x + 2) dy d2 y = 3θ y (3 x + 2)2 2 = 32 θ (θ − 1) y (3.126) dx dx The given equation becomes so that (3 x + 2)

⎛ ez − 2⎞ ⎛ ez − 2⎞ [32 θ (θ − 1) + 3 ⋅ 3θ − 36] y = 3 ⎜ + 4⎜ ⎟ ⎝ 3 ⎠ ⎝ 3 ⎟⎠

(3.127)

1 2z (e − 4) 27 To find the complementary function, we have to solve (θ 2 − 4) y = 0. The auxiliary equation is (θ 2 − 4) y =

m2 − 4 = 0

⇒

m = ±2

yc = c1e 2 z + c2 e −2 z = c1 (3 x + 2)2 + c2 (3 x + 2) −2 To find the particular integral, 1 1 1 ⎛ 1 1 ⎞ ⋅ 2 (e 2 z − 4) = ⎜ z ⋅ e 2 z + ⋅ 4⎟ 27 θ − 4 27 ⎝ 2θ 4 ⎠ 2z 2 1 ⎛ ze ⎞ (3 x + 2) log(3 x + 2) 1 = ⎜ − 1⎟ = + ⎠ 27 ⎝ 4 108 27

yp =

The general solution is y = c1 (3 x + 2)2 + c2 (3 x + 2) −2 +

Ch03.indd 72

(3 x + 2)2 1 log(3 x + 2) + . 108 27

12/14/2011 11:33:17 AM

Linear Differential Equations with Constant Coefficients

3-73

d2 y dy − 6(1 + 2 x ) + 16 y = 8(1 + 2 x )2 . dx 2 dx Solution This is Legendre’s linear equation Put (1 + 2 x ) = e z or z = log(1 + 2 x ) so that

Example 3.2.43 Solve (1 + 2 x )2

dy = 2θ y, dx d2 y (1 + 2 x )2 2 = 2 2 θ (θ − 1) y dx The given equation becomes (1 + 2 x )

[2 2 θ (θ − 1) − 6 ⋅ 2θ + 16] y = 8e 2 z (θ 2 − 4θ + 4) y = 2 x 2 z ⇒ (θ − 2)2 y = 2e 2 z To find the complementary function, we have to solve (θ − 2)2 y = 0. The auxiliary equation is ( m − 2)2 = 0 ⇒ m = 2,2 yc = (c1 + c2 z )e 2 z = [c1 + c2 log(1 + 2 x )](1 + 2 x )2 To find the particular integral, 1 1 yp = = e2 z z 2 2e 2 z = e 2 z 2 (θ − 2) (θ + 2 − 2)2 = (1 + 2 x )2 [log(1 + 2 x )]2 The general solution is y = [c1 + c2 log(1 + 2 x )](1 + 2 x )2 + (1 + 2 x )2 [log(1 + 2 x )]2 .

EXERCISE 3.11 Solve the following: 1. [(5 + 2 x )2 D 2 − 6(5 + 2 x ) D + 8] y = 0. Ans: y = [c1 cosh{ 2 log(5 + 2 x )} + c2 sinh[ 2 log(5 + 2 x )]] ⋅ (5 + 2 x )2

Ch03.indd 73

12/14/2011 11:33:17 AM

3-74

Differential Equations

2. [(2 x + 1)2 D 2 − 2(2 x + 1) D − 12] y = 6 x. Ans: y = c1 (2 x + 1)2 +

c2 3 1 − (2 x + 1) + 2 x + 1 16 4

3. [( x + 2)2 D 2 − ( x + 2) D + 1] y = 3 x + 4. 3 Ans: y = c1 + c2 log( x + 2) ⋅ ( x + 2) + ( x + 2) 2 4. [(1 + x )2 D 2 + (1 + x ) D = 1] y = 4 cos[log(1 + x )]. Ans: y = c1 cos log(1 + x ) + c2 sin log(1 + x ) + 2log(1 + x )sin log(1 + x ) 5. [(1 + x )2 D 2 + (1 + x ) D + 1] y = sin 2[log(1 + x )].

[JNTU 1996]

Ans: y = c1 cos[log(1 + x )] + c2 sin[log(1 + x )] − 13 sin[2(log(1 + x ))]

3.2.4 Method of Variation of Parameters We will now consider the method of variation of parameters for finding the particular integral from the complementary function and hence the general solution of a linear non-homogeneous equation y ′′ + Py ′ + Qy = R where P, Q, R are functions of x. yc = C1u( x ) + C2 v( x ) Let

(3.128) (3.129)

be the company function. Since u and v are solutions of the homogeneous equation y ′′ + Py ′ + Qy = 0

(3.130)

u ′′ + Pu ′ + Qu = 0

(3.131)

v ′′ + Pv ′ + Qv = 0

(3.132)

We have

We assume that a particular integral y p of Eq. (3.128) is given by y p = Au + Bv

Ch03.indd 74

(3.133)

12/14/2011 11:33:17 AM

Linear Differential Equations with Constant Coefficients

3-75

where A and B are now considered as functions of x to be determined. Differentiating Eq. (3.133) with respect to x y ′p = Au ′ + Bv ′ + u

dA dB +v dx dx

Now we choose A and B such that dA dB +v =0 dx dx

(3.134)

y ′p = Au ′ + Bv ′

(3.135)

u Then

Differentiating Eq. (3.135) with respect to x again dA dB (3.136) + v′ dx dx Since y p satisfies Eq. (3.128) we have from Eqs. (3.130), (3.131) and (3.132) y ′′p = Au ′′ + Bv ′′ + u ′

A(u ′′ + Pu ′ + Qu ) + B( v ′′ + Pv ′ + Qv ) + u ′ ⇒

u′

dA dB + v′ =R dx dx

dA dB + v′ =R dx dx

by Eqs. (3.131) and (3.132). Since u( x ) and v( x ) are linearly independent solutions over the interval I we have W (u, v )( x ) =

u v ≠0 u′ v′

(3.137)

Hence, solving Eqs. (3.134) and (3.135) we have, dA dx dB dx −1 = = vR W − uR ⇒ A = −∫

vR uR dx, B = ∫ dx W W

(3.138) (3.139)

Substituting these values in Eq. (3.133) we get the particular integral y p . From Eqs. (3.129) and (3.133) we obtain the general solution or complete solution.

Ch03.indd 75

12/14/2011 11:33:18 AM

3-76

Differential Equations

Note 3.2.44 Since the arbitrary constants C1 and C2 are replaced by A and B in the complementary function and are considered as functions of x for finding the particular integral this method is called “variation of parameters.” Note 3.2.45 The above method is applicable when the coefficients are functions of x or constants. Example 3.2.46 Solve the following equations by the method of variation of parameters. 2 (1) d y + y = cosec x; dx 2 (2) ( D 2 + 1) y = x sin x;

(3) ( D 2 + a 2 ) y = tan ax.

[JNTU 2003]

Solution (1) The given equation is ( D 2 + 1) y = cosec x

(3.140)

P = 0, Q = 1, R = cosec x To find the complementary function, we have to solve ( D 2 + 1) y = 0 To auxiliary equation is m2 + 1 = 0 ⇒ m = ± i

yc = c1 cos x + c2 sin x

(3.141)

We take y = A cos x + B sin x, where A, B are functions of x and u( x ) = cos x, v( x ) = sin x W (u, v )( x ) = A = −∫ = −∫ A= ∫

Ch03.indd 76

cos x sin x =1≠ 0 − sin x cos x

(3.142)

vR dx W

(3.143)

sin xcosec x dx = − x 1

(3.144)

cos xcosec x uR dx = ∫ dx = logsin x W 1

(3.145)

12/14/2011 11:33:18 AM

Linear Differential Equations with Constant Coefficients

3-77

The complete solution is y = c1 cos x + c2 sin x − x cos x + sin x logsin x.

(3.146)

(2) The given equation is ( D 2 + 1) y = x sin x

(3.147)

P = 0, Q = 1, R = x sin x To find the complementary function, we have to solve ( D 2 + 1) y = 0

(3.148)

The auxiliary equation is m2 + 1 = 0 ⇒ m = ± i

yc = c1 cos x + c2 sin x

(3.149)

We take y = A cos x + B sin x, where A, B are functions of x and u( x ) = cos x, v( x ) = sin x W (u, v )( x ) =

cos x sin x = 1± 0 − sin x cos x

(3.150)

vR sin x( x sin x ) −1 1 dx = − ∫ dx = ∫ xdx + ∫ x cos 2 xdx W 1 2 2 x2 sin 2 x cos 2 x =− +x + 4 4 8

A = −∫

cos x( x sin x ) 1 uR dx = − ∫ dx = ∫ x sin 2 xdx 1 2 W x ⎛ cos 2 x ⎞ 1 ⎛ sin 2 x ⎞ − x cos 2 x 1 = ⎜− + sin 2 x ⎟ − ⎜− ⎟= 2⎝ 2 ⎠ 2⎝ 4 ⎠ 4 8

B=∫

The complete solution is 1 ⎛ x2 x ⎞ y = c1 cos x + c2 sin x + ⎜ − + sin 2 x + cos 2 x ⎟ cos x ⎝ 4 4 ⎠ 8 x 1 ⎛ ⎞ + ⎜ − cos 2 x + sin 2 x ⎟ sin x ⎝ 4 ⎠ 8 (3.151)

Ch03.indd 77

12/14/2011 11:33:19 AM

3-78

Differential Equations

y = c1 cos x + c2 sin x x ⎛ x2 ⎞ + ⎜ − cos x − sin x⎟ . ⎝ 4 ⎠ 4 ⎛1 ⎞ ⎜⎝ cos x is merged with the complementary function.⎟⎠ 8

(3.152) (3.153) (3.154)

(3) The given equation is ( D 2 + a 2 ) y = tan ax

(3.155)

P = 0, Q = a 2 , R = tan ax To find the complementary function, we have to solve ( D 2 + a2 ) y = 0 To auxiliary equation is m2 + a 2 = 0 ⇒

m = ± ia

yc = c1 cos ax + c2 sin ax

(3.156)

Let y = A cos ax + B sin ax where A, B are functions of x and u( x ) = cos ax, W (u, v )( x ) =

v( x ) = sin ax

cos ax sin ax =a≠0 − a sin ax a cos ax

(3.157)

sin ax tan ax sin 2 ax vR dx = − ∫ dx = − ∫ dx W a a cos ax 2 1 − cos ax 1 1 = −∫ dx = ∫ cos axdx − ∫ sec axdx a cos ax a a 1 1 = 2 sin ax − 2 log(sec ax + tan ax ) a a

A = −∫

uR cos ax tan ax 1 1 dx = − ∫ dx = ∫ sin ax dx = − 2 cos ax W a a a The complete solution is 1 y = c1 cos ax + c2 sin ax − 2 a B=∫

cos ax log(sec ax + tan ax ).

Ch03.indd 78

(3.158)

12/14/2011 11:33:19 AM

Linear Differential Equations with Constant Coefficients

Example 3.2.47 Solve the equation

3-79

d2 y + y = sec x. dx 2 [JNTU 1995, 1996, 1998, 1999]

Solution The given equation is ( D 2 + 1) y = sec x

(3.159)

The complementary function is yc = c1 cos x + c2 sin x; P = 0, Q = 1, R = sec x Let y = A cos x + B sin x, where A, B are functions of x and u( x ) = cos x, v( x ) = sin x W (u, v )( x ) =

cos x sin x =1≠ 0 − sin x cos x

sin x sec x vR dx = − ∫ dx = log cos x 1 W cos x sec x uR B=∫ dx = − ∫ dx = x 1 W A = −∫

The complete solution is y = c1 cos x + c2 sin x + cos x log x + x sin x. Example 3.2.48 Solve the equation ( D 2 − 2 D ) y = e x sin x. Solution To find the complementary function, we have to solve ( D 2 − 2 D ) y = 0. To auxiliary equation is m2 − 2 m = 0 ⇒

m = 0, m = 2;

P = −2, Q = 0, R = e x sin x yc = c1 + c2 e 2 x u( x ) = 1,

Ch03.indd 79

v( x ) = e 2 x

12/14/2011 11:33:20 AM

3-80

Differential Equations

Let y = A + Be 2 x where A, B are functions of x. W (u, v )( x ) =

1 e2 x = 2e 2 x ≠ 0 0 2e 2 x

vR e 2 x e x sin x dx = − ∫ dx W 2e 2 x 1 ex = − ∫ e x sin x dx = − (sin x − cos x ) 2 4

A = −∫

uR 1.e x sin x 1 dx = ∫ dx = ∫ e − x sin xdx 2x W 2e 2 e− x =− (sin x − cos x ) 4

B=∫

The complete solution is ⎛ ex ⎞ y = c1 + c2 e 2 x + ⎜ − ⎟ (sin x − cos x ) ⋅ ⎝ 4⎠ e− x ex 1+ ( − sin x − cos x )e 2 x y = c1 + c2 e 2 x − sin x. 4 2 Example 3.2.49 Solve the equation ( D 2 + 3D + 2) y = e x + x 2 . Solution Here P = 3, Q = 2, R = e x + x 2 The auxiliary equation is m2 + 3m + 2 = 0 ⇒ m = −1, −2 The complementary function is yc = c1e − x + c2 e −2 x ;

u( x ) = e − x , v( x ) = e −2 x

Let y = Ae − x + Be −2 x where A, B are functions of x. W (u, v )( x ) =

e− x −e − x

e −2 x = −e −3 x ≠ 0 −2e −2 x

vR e −2 x (e x + x 2 ) dx = − ∫ dx W − e −3 x 1 = ∫ e 2 x dx + ∫ x 2 e x dx = e 2 x + e x ( x 2 − 2 x + 2) 2

A = −∫

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Linear Differential Equations with Constant Coefficients

3-81

uR e − x (e x + x 2 ) dx = ∫ dx = − ∫ e 3 x dx W − e −3 x 2 2⎞ −1 ⎛1 = − ∫ x 2 e 2 x dx = e3 x − e x ⎜ x 2 − x + ⎟ ⎝2 3 4 8⎠

B=∫

The complete solution is 1 ⎛ x2 3 ⎞ y = c1e − x + c2 e −2 x + e x + e − x ⎜ − x ⎟ . ⎝ 2 2 ⎠ 6 ⎛ 7 −x ⎞ ⎜⎝ e is merged in the complementary function.⎟⎠ 4

EXERCISE 3.12 Solve the equations by the method of variation of parameters: 1. ( D 2 + 1) y = cosec x cot x. Ans: y = c1 cos x + c2 sin x − cos x log(sin x ) − x sin x 2. ( D + 1)2 y = e − x log x. Ans: y = c1e − x + c2 xe − x +

x 2e − x (1 − 2log x ) + xe −2 x (log x − 1) 4

3. ( D 2 + a 2 ) y = a 2 sec ax. Ans: y = c1 cos ax + c2 sin ax + ax sin ax + cos ax log(cos ax ) 4. ( D 2 − 3D + 2) y = xe x + 2 x. Ans: y = c1e x + c2 e 2 x − 5. ( D 2 + 4) y = tan 2 x.

x2 x 3 e − xe − x + x + 2 2 [JNTU 2003, 2002, 1997, 1994]

Ans: y = c 1 cos 2 x + c2 sin 2 x −

cos 2 x log(sec 2 x + tan 2 x ) 4

6. (( x − 1) D 2 − xD + 1) y = ( x − 1)2 . [Hint: e x , x are basis solutions.] Ans: y = c1e x + c2 x − 1(1 + x + x 2 )

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3-82

Differential Equations

e3 x . x2 Ans: y = c1e 3 x + c2 xe 3 x − e 3 x log x

7. ( D 2 − 6 D + 9) y =

8. ( D 3 + D ) y = cosec x. Ans: y = c1 + c2 cos x + c3 sin x − log(cosec x + cot x ) − cos x logsin x − x sin x 9. ( D 2 − 2 D + 2) y = e x tan x. Ans: y = e x (c1 cos x + c2 sin x ) − e x cos x log(sec x + tan x )

3.2.5

Systems of Simultaneous Linear Differential Equations with Constant Coefficients

In many applied mathematical problems, we encounter systems of simultaneous linear differential equations involving two (or more) dependent variables x and y and one independent variable t, as follows: f1 ( D ) x + f 2 ( D ) y = f (t ) ⎫ g1 ( D ) x + g2 ( D ) y = g (t )⎬⎭

(3.160)

d Where f1 , f 2 , g1 , g2 are rational integral functions of D = with dt constant coefficients and f, g are given functions of t. The system Eq. (3.160) is solved by elimination and we get g 2 ( D ) f (t ) − f 2 ( D ) g (t ) , Δ f ( D ) g (t ) − g1 ( D ) f (t ) y= 1 Δ

x=

where Δ = f1 ( D ) g2 ( D ) − f 2 ( D ) g1 ( D ) ≠ 0. dx dy = 3 x + 2 y, = −5 x − 3 y. dt dt Solution In operator notation, the equations are

Example 3.2.50 Solve

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( D − 3) x − 2 y = 0

(3.161)

5 x + ( D + 3) y = 0

(3.162)

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Linear Differential Equations with Constant Coefficients

3-83

Multiplying Eq. (3.161) by D + 3 and Eq. (3.162) by −2, we get ( D + 3)( D − 3) x − ( D − 3)2 y = 0 − 10 x − ( D + 3)2 y = 0 On subtraction, ( D 2 + 1) x = 0 The auxiliary equation is m2 + 1 = 0 ⇒ m = ± i x = c1 cos t + c2 sin t Substituting in Eq. (3.161) 1 y = ( D − 3)(c1 cos t + c2 sin t ) 2 1 = [ −(c1 sin t + c2 cos t ) − 3c1 cos t − 3c2 sin t ] 2 c − 3c1 , = A cos t + B sin t A = 2 2 c + 3c2 B =1− 1 . 2 Example 3.2.51 Solve

dx dy + 5 x − 2 y = t , + 2 x + y = 0. dt dt

x = 0, y = 0 when t = 0 Solution The given system of equations is

Δ=

( D + 5) x − 2 y = t

(3.163)

2 x + ( D + 1) y = 0

(3.164)

D + 5 −2 = D 2 + 6D + 9 2 D +1

= ( D + 3)2

(3.165) (3.166)

To obtain Δ x , Δ y respectively replace c1 , c2 in Δ by ⎡ f (t ) ⎤ ⎡t ⎤ = ⎣⎢ g (t ) ⎦⎥ ⎣⎢0 ⎦⎥

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3-84

Differential Equations

Δx =

t −2 = 1+ t 0 D +1

(3.167)

Δy =

D+5 t = −2t 2 0

(3.168)

( D + 3)2 x = 1 + t

(3.169)

( D + 3)2 y = −2t

(3.170)

The auxiliary equation is ( m + 3)2 = 0 ⇒ m = −3, −3

(repeated)

The complimentary function is yc = (c1 + c2t )e −3t The particular integral is 1 ( D + 3)2 −2 D⎞ 2⎛ t = − ⎜1 + ⎟ 9⎝ 3⎠ 2 ⎛ 2D t = − ⎜1 − + 9⎝ 3

y p = −2

2 4 ⎞ ⎟⎠ t = − t + 9 27

2 4 ∴ y = (c1 + c2t )e −3t − t + 9 27 Substituting in Eq. (3.164) 2 4 2 x = −( D + 1) y = − y − Dy = −(c1 + c2t )e −3t + t − 9 27 2 −3t − (0 + c2 ⋅ 1 − 3c1 − 3c2t )e + 9 c2 1 ⎛ ⎞ −3t 1 x = ⎜ c1 − + c2t ⎟ e + t + ⎝ ⎠ 2 9 27 t=0 ⇒ x= y=0 c2 1 4 0 = c1 − + 0 = c1 + 2 27 27 4 2 8 2 2 c1 = − c2 = 2c1 + +− + =− 27 27 27 27 9

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Linear Differential Equations with Constant Coefficients

3-85

The required solution is 1 1 (1 + 6t )e −3t + (1 + 3t ) 27 27 2 1 y = − (2 + 3t ) + (2 − 3t ). 27 3 x=−

dx dy = 5 x + y , = y − 4 x. dt dt Solution The given system of equation is Example 3.2.52 Solve

( D − 5) x − y = 0

(3.171)

4 x + ( D − 1) y = 0

(3.172)

Operate Eq. (3.171) by ( D − 1) and add to Eq. (3.172) [( D − 1)( D − 5) + 4]x = 0 ⇒ ( D − 3)2 x = 0 x = (c1 + c2t )e 3t

(3.173)

From Eq. (1) y = −5 x + Dx = −5(c1 + c2t )e 3t + (0 + c2 + 3c1 + 3c2t )e 3t (3.174) y = ( −2c1 + c2 − 2c2t )e 3t .

(3.175)

Example 3.2.53 Solve D 2 x + y = sin t .

(3.176)

x + D 2 y = cos t .

(3.177)

2 Solution Applying D on Eq. (3.176)

D 4 x + D 2 y = − sin t ⇒ ( D 4 − 1) x = − cos t + sin t

(3.178)

The roots of the auxiliary equations are m = 1, −1, i , −i

[by Eq. (3.179)]

xc = c1e t + c2 e − t + c3 cos t + c4 sin t 1 xp = − 2 (cos t + sin t ) ( D − 1)( D 2 + 1) −1 t t = − ⋅ [cos(t + π 2) + sin(t + π 2)] = (sin t − cos t ) −2 2 4

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3-86

Differential Equations

y = sin t − D 2 x = sin t − c1et − c2 e − t + c3 cos t 1 1 t t + c4 sin t − cos t − sin t + sin t − cos t 2 2 4 4 1 = − c1e t − c2 e − t + c3 cos t + c4 sin t + (2 + t )(sin t − cos t ). 4 dx + y = et . dt

(3.179)

dy − x = e −t . dt

(3.180)

Example 3.2.54 Solve

Solution

d 2 x dy + = et dt 2 dt

⇒ ( D 2 + 1) x = et − e − t

xc = c1 cos t + c2 sin t ⎫ ⎪ t −1 1 e − e ⎬ t −t xp = 2 (e − e ) = D +1 2 ⎪⎭ x = c1 cos t + c2 sin t + sinh t y = et − Dx = et + c1 sin t − c2 cos t − y = c1 sin t − c2 cos t + sinh t .

(3.181)

(3.182) et + e − t 2 (3.183)

EXERCISE 3.13 Solve the following pairs of simultaneous linear equations: 1.

dx dy = x − 2 y , = 5 x + 3 y. dt dt Ans: x = (c1 cos6t + c2 sin 6t )e 2t y = [(6c1 − c2 )sin 6t − (c1 + 6c2 ) cos6t ]

2.

e 2t 2

dx dy = − ay, = ax. dt dt Ans: x = c1 cos at − c2 sin at y = c1 sin at − c2 cos at

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Linear Differential Equations with Constant Coefficients

3-87

3. ( D + 2) x + ( D + 1) y = t ; 5 x + ( D + 3) y = t 2 . c1 − 3c2 3c + 3c2 sin t − 1 cos t − t 2 + t + 3 5 5 y = c1 cos t + c2 sin t + 2t 2 − 3t − 4

Ans: x =

4.

dx dy = 3 x + 8 y; = − x − 3 y; dt dt with x(0) = 6, y(0) = −2. Ans: x = 4e t + 2e − t ; y = − et − e − t

5.

dx dy = 2 y − 1, = 1 + 2 x. dt dt

1 Ans: x = c1e 2t + c2 e −2t − ; 2 1 −2 t 2t y = c1e − c2 e + 2 6. ( D + 6) y − Dx = 0; (3 − D ) x − 2 Dy = 0; with x = 2, y = 3 when t = 0. Ans: x = 4e 2t − 2e −3t ; y = e 2t + 2e −3t 7. ( D + 2) x + ( D − 1) y = − sin t ; ( D − 3) x + ( D + 2) y = 4 cos t . 3 2 1 Ans: x = c1e − t s + sin t − cos t ; 5 5 5 y = c1e − t s + sin t + cos t

8. D 2 y = x − 2, D 2 x = y + 2. Ans: x = c1 sin t + c2 cos t + c3e t + c4 e − t + 2; y = − c1 sin t − c2 cos t + c3e t + c4 e − t − 2 9. ( D + 1) x + ( D − 1) y = e t ; ( D 2 + D + 1) x + ( D 2 − D + 1) y = t 2 . 1 1 Ans: x = t 2 − 1 + et ; 2 2 1 2 3 y = t + t − et 2 2

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4 Differential Equations of the First Order but not of the First Degree We will now discuss the solutions of differential equations which are of the first order but are of degree higher than one. Such differential but equations will contain only the first order differential coefficient dy dx dy will occur in a degree higher than one. It is usual to denote dx by p. The general form of such a differential equation is then p n + A1 p n−1 + A2 p n−2 + … + An−1 p + An = 0 where A1, A2 … An are some functions of x and y. Now we will consider various methods of solving the differential equations of the above type.

4.1

EQUATIONS SOLVABLE FOR p

Suppose the differential equation of first order and of degree n > 1 can be solved for p. That is, it can be resolved into n linear factors in p of the type [p − f, (x, y)] [p − f2 (x, y)] … [p − fn (x, y)] = 0 We can equate each factor to zero and the resulting differential equations of the first order and first degree can be solved. Let their solutions be φ1 (x, y, c1) = 0, φ2 (x, y, c2) = 0, … φn (x, y, cn) = 0

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4-2

Differential Equations

where c1, c2,…cn are arbitrary constants. Without loss of generality we can replace the arbitrary constants c1, c2,…cn by a single arbitrary constant c because in any of the above n solutions c is free to take any real value. Thus the n solutions of the given differential equation are φ1 (x, y, c) = 0, φ2 (x, y, c) = 0, …., φn (x, y, c) = 0 Combining the above equations we get a single composite solution as φ1 (x, y, c) = φ2 (x, y, c) = …. φn (x, y, c) = 0 Example 4.1.1 Solve p2 − 5p + 6 = 0. Solution Resolving into linear factors the given differential equation can be written as (p − 3) (p − 2) = 0 Its component equations are p = 3, p = 2. dy Solving p = 3 i.e. = 3 we get y = 3x + c. Also, the solution of dx dy p = 2 i.e. = 2 is y = 2x + c. dx So, the solutions of the given differential equation are y = 3x + c, y = 2x + c The single combined solution is (y − 3x − c) (y − 2x − c) = 0

„

Example 4.1.2 Solve p (p − y) = x (x + y). Solution The given differential equation is p (p − y) = x (x + y) 2 − x − py − xy = 0 p (p + x) (p − x) − y (p + x) = 0 (p + x) (p − x − y) = 0 dy dy = − x, =–y= x dx dx c 1 The first equation gives y = – x 2 + 2 2 2

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Differential Equations of the First Order but not of the First Degree

4-3

The second equation is a linear equation Integrating factor = e−x Its solution is ye−x = c + ∫xe−x dx = c − xe−x − e−x y = cex − x − 1

or

The single combined solution is (2y + x2 − c) (y + x + 1 − cex) = 0

„

Example 4.1.3 Solve 2p2 + p − 1 = 0. Solution 2p2 + p − 1 = (2p − 1) (p + 1) 2p − 1 = 0, dy 2 = 1, dx

p+1=0 dy = –1 dx

Integrating we get 2y = x + c,

y = −x + c

combined solution is (2y − x − c) (y + x − c) = 0

„

Example 4.1.4 Solve p2 − 5px + 6x2 = 0. Solution p2 − 5px + 6x2 = (p − 3x) (p − 2x) = 0 p = 3x, dy = 3 x, dx

p = 2x dy = 2x dx

Integrating we get y=

3 2 x + c, 2

y = x2 + c

combined solution is 3 2 ⎛ ⎞ ⎜⎝ y – x – c⎟⎠ 2

Ch04.indd 3

(y – x

2

)

–c = 0

„

12/9/2011 11:48:59 AM

4-4

Differential Equations

Example 4.1.5 Solve p2 − 5px − 6x2 = 0. Solution p2 − 5px − 6x2 = (p − 6x) (p + x) = 0 p = 6x, dy = 6 x, dx

p = −x dy =–x dx

Integrating we get y=–

y = 3x2 + c,

1 2 x +c 2

combined solution is (y − 3x2 − c)

1 2 ⎛ ⎜⎝ y + x – 2

⎞ c⎟ = 0 ⎠

„

Example 4.1.6 Solve (2p − 3)2 − (p − 2)2 = 0. Solution (2p − 3)2 − (p − 2)2 = (3p − 5) (p − 1) = 0 3p − 5 = 0, dy 3 = 5, dx

p −1 = 0 dy =1 dx

3y = 5x + c,

y=x+c

Integrating we get

combined solution is (3y − 5x − c) (y − x − c) = 0

„

Example 4.1.7 Solve (3p − x)2 − (p − 2x)2 = 0. Solution (3p − x)2 − (p − 2x)2 = (4p − 3x) (2p + x) = 0 4p − 3x = 0, 2p + x = 0 dy dy 4 = 3 x, 2 =–x dx dx Integrating we get 4y =

Ch04.indd 4

3 2 1 x + c , 2 y = – x2 + c 2 2

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Differential Equations of the First Order but not of the First Degree

4-5

combined solution is 3 2 1 2 ⎛ ⎞ ⎛ ⎞ ⎜4y – x – c⎟ ⎜2y + x – c⎟ = 0 2 2 ⎝ ⎠ ⎝ ⎠

„

Example 4.1.8 Solve p2 + 2xp − 3x2 = 0. Solution p2 + 2xp − 3x2 = (p + x)2 − 4x2 = 0 (p + 3x) (p − x) = 0 dy dy = – 3 x, =x dx dx Integrating we get y=–

3 2 1 x + c , y = x2 + c 2 2

combined solution is 3 2 ⎛ ⎜y + x – 2 ⎝

1 ⎞ ⎛ c ⎟ ⎜ y – x2 – 2 ⎠ ⎝

⎞ c⎟ = 0 ⎠

„

Example 4.1.9 Solve p2 − 2py cosh x + 1 = 0. Solution (p − cosh x)2 = −1 + cosh2x = sinh2x (p − cosh x + sinh x) (p − cosh x − sinh x) = 0 (p − e−x) (p − ex) = 0 dy dy = e– x , = ex dx dx Integrating we get y = −e−x + c,

y = ex + c

combined solution is (y + e−x − c) (y − ex − c) = 0

„

Example 4.1.10 Solve yp2 + (x − y) p − x = 0. Solution yp2 + (x − y) p − x = yp (p −1) + x (p − 1) = 0 (p −1) = 0,

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yp + x = 0

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4-6

Differential Equations

dy = 1, dx

y

dy +x= 0 dx

Integrating we get y = x + c, y2 + x2 − c2 = 0 combined solution is (y − x − c) (y2 + x2 − c2) = 0

„

Example 4.1.11 Solve x + yp2 = p (1 + xy). Solution x + yp2 = p (1 + xy) x − xpy = p − yp2 x (1 − yp) = p (1 − yp) (p − x) (yp − 1) = 0 dy dy = x, y =1 dx dx Integrating we get y=

1 2 y2 x + c, = x+ c 2 2

combined solution is 1 2 ⎛ ⎜y – x – 2 ⎝

⎞ ⎛1 ⎞ c ⎟ ⎜ y2 – x – c ⎟ = 0 2 ⎠ ⎝ ⎠

„

Example 4.1.12 Solve xp2 + (y − x) p − y = 0. Solution xp2 + (y − x) p − y = 0 xp2 − xp + yp − y = 0 (p −1) (xp + y) = 0 dy dy = 1, x +y=0 dx dx Integrating we get y = x + c,

xy + c = 0

combined solution is (y − x − c) (xy + c) = 0

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„

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Differential Equations of the First Order but not of the First Degree

4-7

Example 4.1.13 Solve p3 − ax4 = 0. 1

4

Solution p3 − ax4 = 0 ⇒ p = a 3 x 3 Integrating we get 3 73 x 7 7 1 7 ( y + c) = 3 a 3 x 3 1

y + c = a3

343 (y + c)3 = 27 ax7

„

Example 4.1.14 Solve p + px + py + xy = 0. 2

Solution p2 + px + py + xy = 0 p(p + x) + y (p + x) = 0 (p + y) (p + x) = 0 dy dy = – y, –x dx dx Integrating we get log y = – x + c,

1 y = − x2 + c 2

combined solution is

( log y + x – c ) ⎛⎜ y + ⎝

1 2 ⎞ x – c⎟ = 0 2 ⎠

„

Example 4.1.15 Solve (p + y + x) (xp + y + x) (p + 2x) = 0. Solution (p + y + x) (xp + y + x) (p + 2x) = 0 ⇒

dy dy 1 dy + y = –x , + y = –1 , = – 2x dx dx x dx

dy + y = – x linear equation; integrating factor is ex dx Solution is yex = c + ∫ − x ex dx i.e., i.e., x

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yex − c − xex + ex y = ce−x − x + 1

dy d 1 + y= –x ⇒ ( xy ) = – x integrating xy = – x 2 + c dx dx 2

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4-8

Differential Equations

dy = – 2 x integrating y = – x 2 + c dx combined solution is

( y + x –1– ce ) ⎛⎜⎝ xy + 12 x –x

(

Example 4.1.16 Solve y – 1 + p 2

(

Solution y – 1 + p 2

)

−

1 2

−

(

)

−

1 2

2

1

2

p=

1 2

„

= b.

= y −b

(1 + p ) = p =

⎞ – c ⎟ ( y + x2 – c ) = 0 ⎠

=b 1 + p2

We have

)

2

( y − b)

1– ( y – b )

2

−1 =

2

y –b

or

1

( y − b)

2

1− ( y – b )

( y − b)

dy = dx

2

2

1– ( y – b )

2

y –b

Separating the variables and integrating

∫ or

y –b 1– ( y – b )

dy = ∫ dx ⇒ 1– ( y – b ) = x + c 2

2

(x + c)2 + (y − b)2 = 1

„

Example 4.1.17 Solve p (p − y) = x (x + y). Solution p (p − y) = x (x + y) p2 − py − x2 − xy = 0 (p − x) (p + x) − (p + x) y = 0 (p + x) (p − x − y) = 0 dy dy =–x – y = x . This is a linear equation dx dx Integrating factor is e−x Solution is ye−x = c + ∫ xe−x dx = c − xe−x − e−x

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Differential Equations of the First Order but not of the First Degree

Solution of

4-9

dy 1 = – x is y = – x 2 + c dx 2

combined solution is 1 2 ⎞ ⎛ x ⎜ y + x – c ⎟ ( y + x + 1 – ce ) = 0 2 ⎝ ⎠

„

Example 4.1.18 Solve p2 + 2py cot x = y2. Solution p2 + 2py cot x = y2 (p + y cot x)2 = y2 + y2 cot2 x = y2 cosec2 x p + y cot x = ± y cosec x The component equations are dy = y ( – cot x + cosec x ) dx dy = − y ( cot x + cosec x ) dx

(4.1) (4.2)

Separating the variables and integrating (4.1) gives x log y – log c = – log sin x + log tan 2 x x sin 2 cos 2 1 ⎛ y⎞ log ⎜ ⎟ = log = log x x ⎝ c⎠ 2sin 2 cos 2 1 + cos x y=

or Similarly (4.2) gives

c 1 + cos x

x⎞ ⎛ log y – log c = – ⎜ log sin x + log tan ⎟ ⎝ 2⎠ x x sin 2x x ⎛ y⎞ log ⎜ ⎟ = − log 2sin cos ⋅ = − log 2sin 2 x ⎝ c⎠ 2 2 cos 2 2 or

y=

c 1 − cos x

Single combined solution is c c ⎛ ⎞⎛ ⎞ ⎜⎝ y – 1 + cos x ⎟⎠ ⎜⎝ y – 1 – cos x ⎟⎠ = 0

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„

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4-10

Differential Equations

Example 4.1.19 Solve 4y2p2 + 2xy (3x + 1) p + 3x3 = 0. Solution 4y2p2 + 2xy (3x + 1) p + 3x3 = 0 4y2p2 + 6x2yp + 2xyp + 3x3 = 0 2yp (2yp + 3x2) + x (2yp + 3x2) = 0 (2yp + 3x2) (2yp + x) = 0

p= –

3x 2 x , p=– 2y 2y

2y dy = −3x2dx

2y dy = − xdx

Integrating y2 + x3 + c = 0

2y2 + x2 + c = 0

combined solution is (y2 + x3 + c) (2y2 + x2 + c) = 0

„

Example 4.1.20 Solve xyp2 + (x2 + xy + y2) p + x2 + xy = 0. Solution xyp2 + x2p + xyp + x2 + y 2p + xy = 0 xp (yp + x) + x (yp + x) + y (yp + x) = 0 (yp + x) (xp + x + y) = 0 y

dy +x=0 dx

x

dy + y +x = 0 dx

Integrating y2 + x2 + c = 0

2xy + x2 + c = 0

combined solution is (y2 + x2 + c) (2xy + x2 + c) = 0

„

EXERCISE 4.1 Solve the following differential equations: 1. p2 − 7p + 12 = 0. Ans: (y − 4x − c) (y − 3x − c) = 0

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Differential Equations of the First Order but not of the First Degree

4-11

2. y2 + xyp − x2p2 = 0. Ans:

(y

2

– cx1+

5

)( y

2

– cx1–

5

)=0

3. p2 − ax3 = 0. Ans: 25 (y + c)2 − 4ax5 = 0 4. xyp2 − (x2 + y2) p + xy = 0. Ans: (x2 − y2 − c) (y − cx) = 0 5. x2p2 + xyp − 6y2 = 0. Ans: (yx3 − c) (y − cx2) = 0 6. xyp2 + p(3x2 − 2y2) − 6xy = 0. Ans: (y − cx2) (y2 + 3x2 − c) = 0 7. x2p2 − 2xyp + y2 = x2y2 + x4. Ans: y2 − x2 sinh2 (x + c) = 0 8. yp2 + (x − y)p − x = 0. Ans: (y − x − c) (x2 + y2 − c2) = 0 9. x2p2 − 2xyp + 2y2 − x2 = 0. ⎛ y⎞ Ans: sin –1 ⎜ ⎟ = ± log (cx ) ⎝ x⎠ 10. x + y p2 = (1 + xy) p. Ans:

4.2

( ye

−x

y ⎛ ⎞ − c ⎜ x 2 − − c⎟ = 0 ⎝ ⎠ 2

)

EQUATIONS SOLVABLE FOR y

Suppose the given differential equation is solvable for y. Then it can be put in the form y = f (x, p) Differentiating (4.3) w.r.t. x and denoting

(4.3) dy by p, we obtain dx

dp ⎞ ⎛ p = φ ⎜ x, p , ⎟ dx ⎠ ⎝

Ch04.indd 11

(4.4)

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4-12

Differential Equations

which is a differential equation in two variables x and p. Suppose it is possible to solve the differential equation (4.4). Let its solution be f (x, p, c) = 0

(4.5)

where c is an arbitrary constant. Eliminating p between (4.3) and (4.5) we get the required solution of (4.3) in the form ψ(x, y, c) = 0. If it is not easily practicable to eliminate p between (4.3) and (4.5) we may solve (4.3) and (4.5) to get x and y in terms of p and c in the form x = f1 (p, c), y = f2 (p, c), which give us the required solution of (4.3) in the form of parametric equations the parameter being p. Example 4.2.1 Solve y = 3x + a log p. Solution y = 3x + a log p Differentiating w.r.t. x p = 3+ a adp = p (p − 3) dx

1 dp p dx

or

a⎛ 1 1⎞ – ⎟ dp = dx ⎜ 3 ⎝ p – 3 p⎠

a ⎡log ( p – 3) – log p ⎤⎦ = x + log c 3⎣ 3x p–3 = ce a p

or

3x 3 = 1– ce a p

or

p=

3 3x

1– ce a

3x ⎛ ⎞ log p = log 3 – log ⎜1– ce a ⎟ ⎝ ⎠ 3x ⎛ ⎞ y = 3x + a log 3 − a log ⎜1– ce a ⎟ ⎝ ⎠ 2 Example 4.2.2 Solve p − py + 1 = 0.

Solution p2 − py + 1 = 0 ⇒ y = p +

„

1 p

Differentiating w.r.t. x ⎛ 1 ⎞ dp p = ⎜1– 2 ⎟ ⎝ p ⎠ dx

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Differential Equations of the First Order but not of the First Degree

4-13

Separating the variables ⎛1 1⎞ dx = ⎜ – 3 ⎟ dp p p ⎝ ⎠ Integrating x + c = log p –

1 2 p2

The eliminant of p between the given equation and this equation is the solution. „ Example 4.2.3 Solve y = x + a tan−1 p. Solution y = x + a tan−1 p Differentiating w.r.t. x p =1+ a

1 dp 1 + p 2 dx

Separating the variables dx =

a

( p –1) ( p 2 +1)

dp

Putting into partial fractions and integrating 1 1 ⎡1 x + c = a ⎢ log ( p –1) – log p 2 + 1 – tan –1 4 2 ⎣2

(

=

(

1 a log 1+ p 2 2

)

1 2

) ( p – 1) – tan −

–1

p

⎤ p⎥ ⎦ „

Example 4.2.4 Solve 3p5 − py + 1 = 0. Solution 3p5 − py + 1 = 0 ⇒ y = 3 p 4 +

1 p

Differentiating w.r.t. x we get ⎛ 1 ⎞ dp p = ⎜12 p 3 – 2 ⎟ p ⎠ dx ⎝ Separating the variables and integrating

1 ⎞ ⎛ x + c = ∫ ⎜12 p 2 – 3 ⎟ dp p ⎠ ⎝

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4-14

Differential Equations

⇒

1 2 p2

x + c = 4 p3 +

The eliminant of p between the given equation and this equation is the solution. „ Example 4.2.5 Solve y = p2x + p. Solution y = p2x + p Differentiating w.r.t. x we get p = p 2 + ( 2 px + 1)

dp dx

Rewriting the equation as dx 2 1 – x= dp 1– p p – p2 This is a linear equation Integrating factor e –2 ∫

1 2 dp = ( p –1) 1– p

The solution is x ( p – 1) = c + ∫ 2

1 2 p –1) dp = c – p + log p 2 ( p– p

multiplying the given equation by (p − 1)2 and using the value of x we get ( p − 1)2 y = p2 (c − p + log p) + p ( p −1)2

„

Example 4.2.6 Solve y = p sin x + cos x. Solution y = p sin x + cos x Differentiating w.r.t. x p = p cos x – sin x + sin x dp 1– cos x – p =1 dx sin x

or

dp dx

dp x – tan p = 1 dx 2

This is a linear equation. Integrating factor is

Ch04.indd 14

e

x – ∫ tan dx 2

= cos 2

x 2

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Differential Equations of the First Order but not of the First Degree

4-15

Solution is p cos 2

x 1+ cos x 1 1 = c +∫ dx = c + x + sin x 2 2 2 2

This with the given relation is the solution.

„

Example 4.2.7 Solve y + px = p2 x4. Solution y + px = p2 x4 Differentiating w.r.t. x we get dp dp = 2 px 4 + 4 p2 x2 dx dx dp x (1– 2 x 3 p ) + 2 p (1– 2 x 3 p ) = 0 dx ⎞ + 2p⎟ = 0 (1– 2 x3 p ) ⎛⎜⎝ x dp dx ⎠ p+ p+x

1– 2 x3 p = 0, x

dp + 2p = 0 dx

Separating the variables of the second equation dp dx +2 =0 p x Integrating we get log p + 2 log x = log c px2 = c Eliminating p between this and 1− 2x3p = 0 we get the general solution as y + cx = c 2. „ Example 4.2.8 Solve y = 2xp + x2p4. Solution y = 2xp + x2p4 Differentiating w.r.t. x we get dp dp + 4 x2 p3 + 2 p4 x dx dx dp dp p + 2 p4 x + 2x + 4 x2 p3 =0 dx dx

p = 2 p + 2x

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4-16

Differential Equations

p (1 + 2 p 3 x ) + 2 x (1 + 2 p 3 x )

dp =0 dx ⎞ (1 + 2 p3 x ) ⎛⎜⎝ p + 2 x dp ⎟=0 dx ⎠ dp p + 2x =0 dx Separating the variables and integrating 2

dp dx c + = 0 log p 2 x = log c ⇒ p 2 = p x x

Eliminating p from this and the given relation. We get

(y–x p ) 2

(y– x c

2 2

4 2

= 4 x2 p2

2 ⎛ c⎞ x2 ) = 4 x2 ⎜ ⎟ ⎝ x⎠

(y–c )

= 4cx

(y–c )

= 4cx

2 2

The general solution is 2 2

„

Example 4.2.9 Solve y + px = p2 x4. Solution y + px = p2 x4

y = − px + p2x4

Differentiating w.r.t. x and denoting

dy by p we get dx

dp dp + 2 px 4 + 4 x2 p3 dx dx dp 2 p – p 2 x3 + x 1– 2 x 3 p ) = 0 ( dx dp 2 p (1– 2 x 3 p ) + x 1– 2 x 3 p ) = 0 ( dx p = −p – x

Ch04.indd 16

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Differential Equations of the First Order but not of the First Degree

4-17

⎞ (1– 2 x p ) ⎛⎜⎝ 2 p + x dp ⎟=0 dx ⎠ 3

2p + x

∴

dp =0 dx

Separating the variables and integrating dp 2 = – dx p x px2 = c

we get

Substituting in the given relation y=–x

c c2 4 + x x2 x4

or xy = c 2 x – c „

which is the required solution. Example 4.2.10 Solve y = 2px − p2. Solution y = 2px − p2 Differentiating w.r.t. x p = 2 p + 2x or

p+2

or or

dp dp – 2p dx dx

dp ( x – p) = 0 dx dx p + 2x = 2 p dp dx 2 + x=2 dp p

This is a linear equation

The solution is or

Ch04.indd 17

∫

2 dp p

= e 2 log p = p 2 2 xp 2 = ∫ 2 p 2 dp + c = p 3 + c 3 2 x = p + cp –2 3

Integrating factor is e

(4.6)

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4-18

Differential Equations

Here it is not easily practicable to eliminate p between this and the given relation. So putting the value of x in the given relation

or

⎛2 ⎞ y = 2 p ⎜ p + cp –2 ⎟ – p 2 ⎝3 ⎠ 1 y = p 2 + 2cp –1 3

Equations (4.6) and (4.7) constitute the solution.

(4.7) „

2

dy ⎛ dy ⎞ +⎜ ⎟ . dx ⎝ dx ⎠ Solution y − x = xp + p2 or y = x (1 + p) + p2 Differentiating w.r.t. x

Example 4.2.11 Solve y – x = x

dp dp + 2p dx dx dp dp 1+ x +2 =0 dx dx dx + x = –2p dp

p =1+ p + x or or

Linear equation with integrating factor e p. Its solution is xe p = c − ∫ 2pe p dp = c − 2 (p − 1) e p or

x = −2( p−1) + ce −p

(4.8)

Substituting this value of x in the given relation y = (1 + p) [ −2 (p − 1) + ce −p] + p2 y = e (1 + p) e −p + 2 − p2

(4.9)

Equations (4.8) and (4.9) constitute the solution.

„

or

Example 4.2.12 Solve y = a 1 + p 2 . Solution y = a 1 + p 2 Differentiating w.r.t. x p=a

Ch04.indd 18

(

1 1 + p2 2

)

−

1 2

2p

dp dx

12/9/2011 11:49:03 AM

Differential Equations of the First Order but not of the First Degree

a dp

or

1 + p2

Integrating we get

4-19

= dx

(

)

x = a log p + 1 + p 2

(4.10)

From the given relation y2 = a2 + a2p2 p=

or putting 1 + p 2 =

y a

1 2 y – a2 a

and p =

1 a

y 2 – a 2 in (4.10)

y⎤ ⎡1 x = a log ⎢ y2 – a2 + ⎥ + c a a ⎣ ⎦

)

(

x = a log y + y 2 – a 2 + c1

or

where c1 = c − a log a Hence the required solution is

x = a log ⎡ y + y 2 – a 2 ⎤ + c ⎣ ⎦

„

EXERCISE 4.2 Solve the following differential equations: 1. y = 2px + p4x2. Ans: (y − c2)2 = 4cx 2. 4p3 + 3xp = y. 3 1 − − 12 2 3 p + cp 2 ; y = – p 3 + 3cp 2 7 7 3. x2 + p2x = yp.

Ans: x = –

(

1

– 13 p 2 + cp 2 1 1 3 2 Ans: x = – p + cp ; y = 3 p

Ch04.indd 19

)

2

1 ⎞ ⎛ 1 + p 2 ⎜ – p 2 + cp 2 ⎟ ⎝ 3 ⎠

12/9/2011 11:49:04 AM

4-20

Differential Equations

4. y − 2px = xp2. Ans: y = 2c x + c 2 5. p2 − xp + y = 0. Ans: y = cx − c2 6. x3p2 + x2yp + 4 = 0. Ans: cxy + 4x + c2 = 0 7. 2xp3 − 6yp2 + x4 = 0. Ans: 2c3x3 = 1 − 6 c2y 8. x − yp = ap2. Ans: y + ap =

4.3

1 1– p

( c + a sin p ) –1

2

EQUATIONS SOLVABLE FOR x

Suppose the given differential equation is solvable for x. Then it can be put in the form x = f (y, p) Differentiating (4.11) w.r.t. y and writing

(4.11) 1 dx for we get p dy

⎛ 1 dp ⎞ = φ ⎜ y , p, ⎟ p dy ⎠ ⎝

(4.12)

which is a differential equation in two variables y and p. Suppose it is possible to solve the differential equation (4.12). Let its solution be f (y, p, c) = 0

(4.13)

where c is the arbitrary constant. Eliminating p between (4.11) and (4.12) we get the required solution of (4.11) in the form x = f1 (p, c),

y = f2 (p, c)

(4.14)

which give us the required solution of (4.11) in the form of parametric equations, the parameter being p.

Ch04.indd 20

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Differential Equations of the First Order but not of the First Degree

4-21

Example 4.3.1 Solve y = 2px + y2p3. Solution The given differential equation can be written as

x=

y y 2 p3 – 2p 2

Differentiating w.r.t. y we get 1 1 y dp 2 yp 2 y 2 dp = – – – 2p 2 p 2 p 2 p dy 2 2 dy

⎛ y ⎞ dp 1 + yp 2 + ⎜ 2 + py 2 ⎟ =0 2p ⎝ 2p ⎠ dy

or

⎛ 1 ⎞ ⎛ 1 ⎞ dp p ⎜ 2 + py ⎟ + y ⎜ 2 + py ⎟ =0 ⎝ 2p ⎠ ⎝ 2p ⎠ dy ⎛ 1 ⎞⎛ dp ⎞ ⎜⎝ 2 p 2 + py ⎟⎠ ⎜⎝ p + y dy ⎟⎠ = 0

or or

dp dp dy + = 0 or py = c = 0 we have dy p y c in the given relation Substituting the value of p i.e. p = y ⎛ c2 ⎞ ⎛c⎞ 2cx c 2 y = 2 x ⎜ ⎟ + y 2 ⎜ 2 ⎟ or y = + y y ⎝ y⎠ ⎝y ⎠

From p + y

or

y2 = 2cx + c2 which is the required solution.

„

Example 4.3.2 Solve p3 − 4xyp + 8y2 = 0. p2 2 y + Solution 4xyp = p3 + 8y2 or x = p 4y Differentiating w.r.t. y we get 1 p2 2 p dp 2 2 y dp =– + + – 2 p 4y 4 y dy p p 2 dy or

Ch04.indd 21

dp ⎛ p 2y ⎞ p2 1 – = – ⎜ 2 ⎟ 2 dy ⎝ 2 y p ⎠ 4 y p

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4-22

Differential Equations

⎛ p p ⎛ p 2y ⎞ 2y ⎞ – 2⎟= – 2⎟ ⎜ ⎜ ⎝ 2y p ⎠ 2y ⎝ 2y p ⎠ dp p p 2y – = 0, – = 0 dy 2 y 2 y p2

dp dy

or or

The first equation gives p2 = cy. From the differential equation, we have 8y2 = p (4xy − p2) Substituting p = cy in this equation 1

1

3

8 y 2 = ( cy ) 2 ( 4 xy – cy ) or 8 y 2 = c 2 y 2 ( 4 x – c ) 1

or

1

8 y 2 = c 2 (4 x – c)

or y=

or 64 y = c ( 4 x – c )

2

1 ⎛ 1 ⎞ 1 2 e ⎜ x – e ⎟ or y = c ( x – c ) writing c for c 4 ⎝ 4 ⎠ 4

Hence the required solution is y = c (x − c)2.

„

Example 4.3.3 Solve ap + py − x = 0. 2

Solution x = py + ap2 Differentiating w.r.t. y and writing

1 dx for p dy

1 dp dp =p+ y + 2 ap p dy dy

1– p 2 dp dp =y + 2ap p dy dy dy p 2ap 2 + 2 y=– 2 dp p – 1 p –1

or or

∫

Linear differential equation. I.F. is e The solution is

(

)

y p2 – 1

Ch04.indd 22

1 2

=

p p 2 –1

dp

1

= ( p –1) 2

(

)

1 p 2 –1 + 1 –2ap 2 2 2 dp + c = – 2a p – 1 ∫ p2 – 1 ∫ p 2 –1 dp + c

(

)

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Differential Equations of the First Order but not of the First Degree

⎛ = – 2a ∫ ⎜ p 2 – 1 + ⎝

⎞ ⎟ dp + c p2 – 1 ⎠ 1

1 ⎡1 = –2a ⎢ p p 2 – 1 – cosh –1 p + cosh –1 2 ⎣2 = – ap or

y=

4-23

⎤ p⎥ + c ⎦

p 2 – 1 – a cosh –1 p + c

c – a cosh –1 p p2 – 1

– ap

(4.15)

Substituting this in the given differential equation we get

⎛ c – a cosh –1 p ⎞ x= p⎜ – ap ⎟ + ap 2 ⎜ ⎟ p 2 –1 ⎝ ⎠ or

x=

p ( c – a cosh –1 p ) p 2 –1

Equations (4.15) and (4.16) constitute the general solution.

(4.16) „

Example 4.3.4 Solve xp3 = a + bp. a b Solution xp3 = a + bp or x = 3 + 2 p p Differentiating w.r.t. y 1 3a dp 2b dp =– 4 – p p dy p 3 dy or

⎛ 3a 2b ⎞ dp 1 = –⎜ 3 + 2 ⎟ p ⎠ dy ⎝p

or

⎛ 3a 2b ⎞ dy = – ⎜ 3 + 2 ⎟ dp p ⎠ ⎝p

Integrating y=

Ch04.indd 23

3a 2b + +c 2 2p p

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4-24

Differential Equations

The general solution is a b x= 3 + 2 p p

and

y=

3a 2b + +c 2 p 2p

„

Example 4.3.5 Solve y2 log y = xyp + p2. y log y p – Solution y2 log y = xyp + p2 or x = p y Differentiating w.r.t. y we get 1 1 y log y dp 1 dp p = (1 + log y ) – – + p p p 2 dy y dy y 2

p ⎛ y log y 1 ⎞ ⎛ y log y 1 ⎞ dp + ⎟ –⎜ + ⎟ =0 y ⎜⎝ p 2 y ⎠ ⎝ p2 y ⎠ dy

⎛ y log y 1 ⎞ ⎛ p dp ⎞ + ⎟⎜ – ⎜ ⎟=0 2 y ⎠ ⎝ y dy ⎠ ⎝ p p dp – = 0 ⇒ p = cy y dy Eliminating p between this and the given relation y2 log y = cxy2 + c2y2 or log y = cx + c2 which is the general solution.

⎛ p ⎞ Example 4.3.6 Solve p = tan ⎜ x – ⎟. + p2 ⎠ 1 ⎝ p Solution Solving for x, x = tan –1 p + 1 + p2 Differentiating w.r.t. y 2 p dp = dy 2 2 1 + p ( )

„

(4.17)

Integrating c–

1 = y 1 + p2

Equations (4.17) and (4.18) constitute the general solution.

Ch04.indd 24

(4.18) „

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Differential Equations of the First Order but not of the First Degree

4-25

Example 4.3.7 Solve ayp2 + (2x − b) p − y = 0. Solution Solving for x, 2 x =

y – ayp + b . p

Differentiating w.r.t. y 2 1 y dp dp = – 2 – ap – ay p p p dy dy

⎛ ⎞ dp ⎞ ⎛ 1 ⎜⎝ p + y dy ⎟⎠ ⎜⎝ p 2 + a⎟⎠ = 0 p+ y

From the first factor

dp =0 dy

py = c

Integrating

Eliminating p between this and the original equation ac 2 ( 2 x – b ) c + – y =0 y y ac2 + (2x − b) c − y2 = 0

i.e.

„

This is the solution. Example 4.3.8 Solve 2px = 2 tan y + p3 cos2 y. Solution Solving for x

x=

tan y p 2 cos 2 y + p 2

Differentiating w.r.t. y 1 1 1 dp dp = sec 2 y – 2 tan y + p cos 2 y – p 2 sin y cos y p p p dy dy

⎛ 1 tan y ⎞ dp tan 2 y – p 2 sin y cos y + ⎜ p cos 2 y – =0 ⎟ p p 2 ⎠ dy ⎝ ⎛ tan y ⎞ ⎛ tan y ⎞ dp − p tan y ⎜ p cos 2 y – + ⎜ p cos 2 y – =0 2 ⎟ p ⎠ ⎝ p 2 ⎟⎠ dy ⎝ ⎛ ⎞ tan y ⎞ ⎛ dp 2 ⎜⎝ p cos y – p 2 ⎟⎠ ⎜⎝ dy – p tan y ⎟⎠ = 0

Ch04.indd 25

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4-26

Differential Equations

dp – p tan y = 0 separating the variables and integrating dy log p = log sec y + log c or p = c sec y

Eliminating p between this and the original equation 2cx sec y = 2 tan y + c3 sec3 y cos2 y 2cx = 2 sin y + c3

or

„

which is the general solution. EXERCISE 4.3 Solve the following differential equations: 1. x = 2 p 3 +

y . p

Ans: x = 2c 3 + 2. p2y + 2px = y.

y c

Ans: y2 = 2cx + c2 3. y = 2px + y2p3. Ans: y2 = 2cx + c2 4. 4xp2 + 4yp − y4 = 0. Ans: 4c (cxy + 1) = y 5. p3 − 4xyp + 8y2 = 0. Ans: c (c − 4x)2 = 64y 6. p3 − (y + 3) p + x = 0. Ans: x = 2 p +

c 2

p –1

; y = p2 – 1 +

c p 2 –1

7. y2 log y = xyp + p2. Ans: log y = cx + c2 8. 4 (xp2 + yp) = y4. Ans: y = 4c (xyc + 1)

Ch04.indd 26

12/9/2011 11:49:07 AM

5 Linear Equation of the Second Order with Variable Coefficients An equation of the form d2 y dy (5.1) + P + Qy = R 2 dx dx where P, Q, R are the real valued functions of x defined on an interval I is called a linear equation of the second order with variable coefficients. There is no known general method to solve such an equation. We discuss below three methods which are useful in the solution of second order linear equations with variable coefficients. 1. Change of the dependent variable when part of the complementary function is known. 2. Change of the dependent variable and removal of the first derivative or reduction to normal form. 3. Change of the independent variable. General solution of Eq. (5.1) in forms of a given or known solution of the complementary function. Let

y = uv

(5.2)

Be the general solution of Eq. (5.1) where u ( x ) is a solution of the reduced equation i.e. d 2u du +P + Qu = 0 2 dx dx

Ch05.indd 1

(5.3)

12/14/2011 11:05:36 AM

5-2

Differential Equations

Differentiating (5.2) we get

dy dv du =u +v dx dx dx

(5.4)

d2 y d 2v du dv d 2u = u + 2 + v dx 2 dx 2 dx dx dx 2

(5.5)

From (5.1), (5.2) and (5.5) we get du dv d 2u ⎞ du ⎞ ⎛ d 2v ⎛ dv + v 2 ⎟ + P ⎜ u + v ⎟ + Quv = R ⎜⎝ u 2 + 2 ⎝ dx dx dx dx dx ⎠ dx ⎠ du dv ⎞ du dv ⎛ d 2u ⎞ ⎛ d 2v v⎜ 2 + P + Qu⎟ + u ⎜ 2 + P ⎟ + 2 =R ⎝ dx ⎠ ⎝ dx dx dx ⎠ dx dx dv ⎞ du dv ⎛ d 2v u⎜ 2 + P ⎟ + 2 = R, ⎝ dx dx ⎠ dx dx u

by (5.3)

d 2v ⎛ du ⎞ dv Pu + 2 ⎟ =R 2 ⎜ dx ⎝ dx ⎠ dx

d 2v ⎛ 2 du ⎞ dv R +⎜P + = ⎟ 2 dx ⎝ u dx ⎠ dx u dV ⎛ 2 du ⎞ R dV +⎜P + V = where V = ⎟ dx ⎝ u dx ⎠ u dx

(5.6) (5.7)

This is a linear equation in V. 2 du ⎞ ⎤ ⎡ ⎛ I.F = exp ⎢ ∫ ⎜ P + ⎟ dx = exp ⎡⎣ ∫ Pdx + 2log u ⎤⎦ ⎝ u dx ⎠ ⎥⎦ ⎣ P dx = u2e∫ The solution of Eq. (5.7) is R P dx P dx V u 2 e ∫ = ∫ u 2 e ∫ dx + c1 u − P dx dv e ∫ ⎡ R u e ∫ Pdx dx + c ⎤ V= = 1⎥ 2 ⎦ dx u ⎢⎣ ∫

Ch05.indd 2

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Linear Equation of the Second Order with Variable Coefficients

5-3

Integrating we get − Pdx ⎡ e − ∫ Pdx ⎤ e ∫ Pdx V = ∫ ⎢ 2 ∫ R u e ∫ dx ⎥ dx + c1 ∫ 2 dx + c2 u ⎢⎣ u ⎥⎦

Substituting this value of v in (5.2) we get − Pdx ⎡ e − ∫ Pdx ⎤ e ∫ Pdx y = c2 u + c1u ∫ 2 dx + u ⎢ ∫ 2 ∫ R u e ∫ ⎥ dx u ⎢⎣ u ⎥⎦

Since this includes the known solution y = u ( x ) and it contains two arbitrary constants, it is the general solution of (5.1).

5.1 TO FIND THE INTEGRAL IN C.F. BY INSPECTION, d 2y dy +p +Qy =0 i.e. TO FIND A SOLUTION OF 2 dx dx d2 y dy + P + Qy = 0 2 dx dx

(5.8)

1. y = e mx is a solution of (5.8) if m2 + pm + Q = 0 we have y = e mx

dy d2 y = me mx , 2 = m2 e mx dx dx

If y = e mx is a solution of (5.8) then ( m2 + P ( m) + Q ) e mx = 0

∴

m2 + Pm + Q = 0

Deductions y = e x is a solution of (5.8) if 1 + P + Q = 0 y = e − x is a solution of (5.8) if 1 − P + Q = 0 2. y = x m is a solution of (5.8) if m ( m − 1) + Pm x + Qx 2 = 0 dy d2 y = mx m −1 , = m ( m − 1) x m − 2 y = xm , dx dx 2 y = x m is a solution of (5.8) then m ( m − 1) x (m − 2) + Pm x (m −1) + Q x m = 0

or

m ( m − 1) + Pm x + Q x 2 = 0

Deductions y = x is a solution of (5.8) if P + Qx = 0 y = x 2 is a solution of (5.8) if 2 + 2 Px + Qx 2 = 0

Ch05.indd 3

12/14/2011 11:05:37 AM

5-4

Differential Equations

The above results can be summarized as follows 1. 2. 3. 4. 5. 6.

y = e x is a part of C.F if 1 + P + Q = 0 y = e – x is a part of C.F if 1 − P + Q = 0 y = e mx is a part of C.F if m2 + mP + Q = 0 y = x is a part of C.F if P + Qx = 0 y = x 2 is a part of C.F if 2 + 2 Px + Qx 2 = 0 y = x m is a part of C.F if m ( m − 1) + Pm x + Qx 2 = 0

Example 5.1.1 Solve

( x + 2)

d2 y dy − ( 2 x + 5) + 2 y = ( x + 1) e x . 2 dx dx

Solution The given solution is

d 2 y 2 x + 5 dy 2 x +1 x − + y= e dx 2 x + 2 dx x + 2 x+2

(5.9)

d2 y dy + p + Qy = R 2 dx dx

Comparing with We have

P=−

2x + 5 2 x +1 x ,Q = ,R= e x+2 x+2 x+2

Here

22 + 2 P + Q = 0

∴

y = e 2 x is a part of the C.F of the Eq. (5.9)

Putting

dy dv = e2 x + 2v e 2 x dx dx d2 y d 2v dv 2x = e + 4e 2 x + 4 ve 2 x in. (5.9) we get 2 2 dx dx dx d 2 v 2 x + 3 dv x + 1 − x + = e dx 2 x + 2 dx x + 2 y = ve 2 x ,

dp 2 x + 3 x + 1 −x + P= e dx x + 2 x+2

or

dv ⎞ ⎛ ⎜⎝ P = ⎟⎠ dx

This is linear in p I.F. = e

Ch05.indd 4

∫

2x +3 dx x+2

⎛

=e

1 ⎞

∫ ⎜⎝ 2 − x + 2 ⎟⎠

dx

=

e2 x x+2

12/14/2011 11:05:39 AM

Linear Equation of the Second Order with Variable Coefficients

5-5

The solution is p

e2 x x + 1 − x e2 x =∫ e dx + c1 x+2 x+2 x+2 1 ⎞ x ⎛ 1 = ∫⎜ − e dx + c1 ⎝ x + 2 x + 22 ⎟⎠ ∫ 1 x = e + c1 x+2

∫ ( f ( x ) + f ′ ( x )) e

x

= f ( x )e x

dv = e − x + e2 e −2 x ( x + 2) dx Integrating this we get

∴ p=

1 1 v = − e − x − c1e −2 x ( x + 2) − c1e −2 x + c2 2 4 1 −x −2 x = − e − c1 ( 2 x + 5) e + c2 4 The general solution of (5.9) is 1 y = ve 2 x = −e x − c1 ( 2 x + 5) + c2 e 2 x 4

„

d2 y dy − ( 2 x − 1) + ( x − 1) y = 0. 2 dx dx Solution Writing the differential equation in the standard form

Example 5.1.2 Solve

x

d2 y ⎛ 1 ⎞ dy ⎛ 1 ⎞ − ⎜2 − ⎟ + ⎜1 − ⎟ y = 0 2 ⎝ dx x ⎠ dx ⎝ x⎠

(5.10)

Here 1 + P + Q = 0, so that y = e x is party of the CF dy dv d 2 y d 2v dv Put y = ve x , = e x + ve x , 2 = 2 e x + 2 e x + ve x in (5.10) dx dx dx dx dx We get

dv d 2 v 1 dv dp 1 + = 0 i.e. + p = 0 where p = 2 dx dx x dx dx x Separating the variables and integrating Log p = − log x + log c1 dv c1 p= = or dx x

Ch05.indd 5

12/14/2011 11:05:40 AM

5-6

Differential Equations

Integrating again v = c1 log x + c2

The complete solution of (5.10) is

y = ve x = (c1 log x + c2 ) e x

„

d2 y dy − ( x 2 + 2 x ) + ( x + 2) y = x 3e x . 2 dx dx Solution Writing the equation in the standard form

Example 5.1.3 Solve

x2

d2 y ⎛ − ⎜1 + dx 2 ⎝

⎛ P + Qx = − ⎜1 + ⎝

Here ∴

2 ⎞ dy ⎛ 1 2 ⎞ + ⎜ + ⎟ y = xe x ⎟ x ⎠ dx ⎝ x x 2 ⎠

(5.11)

2⎞ ⎛ 1 2 ⎞ ⎟ +⎜ + ⎟ x=0 x ⎠ ⎝ x x2 ⎠

y = x is a part of the C. F. of (5.11)

Putting y = vx,

dy dv d2 y d 2v dv in (5.11) = x + v, 2 = x 2 + 2 dx dx dx dx dx

We get

d 2 v dv − = ex dx 2 dx

or

dp − p = ex dx

dv ⎞ ⎛ ⎜⎝ p = ⎟⎠ dx

which is linear in p. I. F. = e ∫

− dx

= e− x

∴ The solution is p e − x = ∫ e x e − x dx + c1 = x + c1

∴

p=

dv = xe x + c1e x dx

Integrating this we get v = xe x − e x + c1e x + c2

The complete solution of (5.11) is y = vx = x 2 e x − xe x + c1 xe x + c2 x

Ch05.indd 6

„

12/14/2011 11:05:40 AM

Linear Equation of the Second Order with Variable Coefficients

5-7

d2 y = 2 y given that y = cot x is a solution. dx 2 dy dv = cot x − v cosec2 x , Solution Put y = v cot x , dx dx d 2 y d 2v dv = 2 cot x − 2 cosec2 x + 2 v cosec2 x cot x 2 dx dx dx

Example 5.1.4 Solve sin 2 x

The given differential equation becomes

d 2v dv −2 =0 2 dx dx 2 d v 2 dv − =0 or 2 dx sin x cos x dx dp 2 dv ⎞ ⎛ − p=0 or ⎜⎝ p = ⎟⎠ dx sin x cos x dx Separating the variables sin 2 x cot x

dp = 4 cosec 2 x dx p Integrating

or ∴

1 log p = 4 log tan x + log c1 2 dv p= = c tan 2 x = c1 (sec2 x − 1) dx 1 v = c1 ( tan x − x ) + c2

The complete solution of the given equation is y = v cot x = c1 (1 − x cot x ) + c2 cot x Example 5.1.5 Solve (1 − x 2 )

„

3 d2 y dy 2 2 . + x − y = x 1 − x ( ) dx 2 dx

Solution Here P + Qx = 0. So y = x is a solution of the equation

(1 − x 2 )

Ch05.indd 7

d2 y dy +x −y=0 2 dx dx

12/14/2011 11:05:41 AM

5-8

Differential Equations

dy dv and =v+x dx dx d2 y d 2v dv = x +2 2 2 dx dx dx

Substituting y = vx and hence

The given differential equation becomes 1 d 2v ⎛ 2 x ⎞ dv +⎜ + = (1 − x 2 ) 2 ⎟ 2 2 dx ⎝ x 1 − x ⎠ dx

Substituting p for

dv this becomes dx

1 dp ⎛ 2 x ⎞ +⎜ + p = (1 − x 2 ) 2 ⎟ 2 dx ⎝ x 1 − x ⎠

x2 This is a linear equation with integrating factor Its solution is x2 1 p = x 3 + c1 1 1 − x2 2 3

(

1

(1 − x 2 ) 2

)

c dv 1 = x 1 − x 2 + 12 1 − x 2 dx 3 x Integrating again we get 3 ⎛ 1 1 − x2 ⎞ +c v = − (1 − x 2 ) 2 + c1 ⎜ sin −1 x + 9 x ⎟⎠ 2 ⎝

or

p=

Hence the complete solution is

(

)

3 1 y = vx = − x (1 − x 2 ) 2 + c1 x sin −1 x + 1 − x 2 + c2 x 2

Example 5.1.6 Solve ( x sin x + cos x )

„

d2 y dy − x cos x + y cos x = 0 2 dx dx

given that y = x is a solution. Solution Substituting y = vx, we get

d 2 v 2 x sin x + 2cos x − x 2 cos x dv + =0 dx 2 x 2 sin x + x cos x dx

Ch05.indd 8

12/14/2011 11:05:41 AM

Linear Equation of the Second Order with Variable Coefficients

5-9

dv , we get dx dp 2 ( x sin x + cos x ) − x 2 cos x + p=0 dx x ( x sin x + cos x )

Substituting p for

dp ⎛ 2 x cos x ⎞ +⎜ − ⎟ p=0 dx ⎝ x x sin x + cos x ⎠

or

dp ⎛ 2 x cos x ⎞ +⎜ − ⎟ dx p ⎝ x x sin x + cos x ⎠ Integrating we get px 2 =c x sin x + cos x dv ⎛ sin x cos x ⎞ p= = c⎜ + 2 ⎟ or ⎝ x dx x ⎠ Integrating again or

cos x sin x ⎞ ⎛ sin x v = c⎜∫ dx − −∫ dx ⎟ + c2 ⎝ x ⎠ x x 1 = c1 cos x + c2 x ∴

y = vx = c1 cos x + c2 x

„

d2 y dy − ( 2 x + 5) + 2 y = ( x + 1) e x . 2 dx dx is a solution of the equation

Example 5.1.7 Solve ( x + 2) Solution Here y = e 2 x

( x + 2)

d2 y dy − ( 2 x + 5) + 2 y = 0 2 dx dx

Substituting y = ve 2 x and hence dy ⎛ dv ⎞ = ⎜ + 2v ⎟ e 2 x ⎝ ⎠ dx dx

and

d 2 y ⎛ d 2v dv ⎞ = ⎜ 2 + 4 + 4v⎟ e 2 x 2 ⎝ ⎠ dx dx dx

The given differential equation becomes

( x + 2)

Ch05.indd 9

d 2v dv + ( 2 x + 3) = ( x + 1) e − x 2 dx dx

12/14/2011 11:05:42 AM

5-10

Differential Equations

Substituting p for

dv this equation becomes dx dp 2 x + 3 ( x + 1) e − x + p= dx x + 2 ( x + 2)

This is a linear equation with the integrating factor e 2 x ( x + 2) ∴

p e 2 x ( x + 2) = ∫ −1

p=

or

x +1

( x + 2)

2

e x dx + c1 =

−1

ex +c ( x + 2) 1

dv = e − x + c1e −2 x ( x + 2) dx

Integrating again

1 v = − e − x − c1 ( 2 x + 5) e 2 x + c2 4 Hence the complete solution is 1 y = v e 2 x = − e x − c1 ( 2 x + 5) + c2 e 2 x 4 Example 5.1.8 Solve

„

d2 y ⎛ 2 2⎞ + ⎜1 + cot x − 2 ⎟ y = x cos x given that 2 ⎝ dx x x ⎠

sin x is a complementary function. x v sin x Solution Substituting y = x dy v1 sin x ⎛ cos x − sin x ⎞ = + v⎜ ⎟⎠ and ⎝ dx x x2 d 2 y d 2 v sin x dv ⎛ cos x sin x ⎞ ⎛ sin x 2cos x 2sin x ⎞ = 2 +2 ⎜ − 2 ⎟ + v⎜− − + ⎟ 2 ⎝ dx dx x dx ⎝ x x ⎠ x x2 x3 ⎠ The given equation reduces to d 2v 1 ⎞ dv ⎛ + 2 ⎜ cot x − ⎟ = x 2 cot x 2 ⎝ dx x ⎠ dx Putting

Ch05.indd 10

dv = v the equation becomes dx dv 1⎞ ⎛ + 2 ⎜ cot x − ⎟ v = x 2 cot x ⎝ dx x⎠

12/14/2011 11:05:42 AM

Linear Equation of the Second Order with Variable Coefficients

This is a linear equation and the integrating factor is ∴ Its solution is

v∫

or or v =

5-11

sin 2 x x2

sin 2 x 1 = ∫ sin 2 x dx + c1 x2 2 1 1 1 = − cos 2 x + c1 = − + sin 2 x + c1 4 4 2 dv 1 1 = − cosec2 2 x + x 2 + c1 x 2 cosec2 x dx 4 2

1 3 ⎛ 1⎞ x + ⎜ c1 − ⎟ ⎡⎣ − x 2 cot x + 2 x log sin x − 2∫ logsin x dx ⎤⎦ + c2 ⎝ 6 4⎠

The complete solution is y=

{

v sin x sin x ⎡ 1 3 ⎛ 1⎞ x + ⎜ c1 − ⎟ − x 2 cot x + 2 x logsin x − 2∫ = ⎢ ⎝ x x ⎣6 4⎠ ⎤ „ ∫ logsin x dx + c2⎥⎦

}

d2 y dy − cot x − (1 − cot x ) y = e x sin x. dx 2 dx Solution Here sum of the coefficients is zero. So, e x is a solution. Substituting y = ve x we get Example 5.1.9 Solve

d 2v dv + ( 2 − cot x ) = sin x 2 dx dx dv = p there the equation becomes dx dp + ( 2 − cot x ) p = sin x dx This is a linear equation. Its integrating factor is

Let

e∫

(2 − cot x )dx

= e 2 x − log sin x =

e2 x sin x

Its solution is p

Ch05.indd 11

e2 x = e 2 x dx + c1 sin x ∫

12/14/2011 11:05:43 AM

5-12

Differential Equations

or ∴

1 p = sin x + c1e −2 x sin x 2 1 v = − cos x + c1 ∫ e −2 x sin x dx + c2 2 1 1 = − cos x − c1e −2 x ( cos x + 2sin x ) + c2 2 5

The complete solution is 1 1 y = v e x = − e x cos x − c1e − x ( cos x + 2sin x ) + c2 e x 2 5

„

Example 5.1.10 Solve xy ′′ − 2 ( x + 1) y ′ + ( x + 2) y = ( x − 2) e 2 x . Solution Dividing by x, the given equation in the standard form is ⎛ 1⎞ ⎛ 2⎞ ⎛ 2⎞ y ′′ − 2 ⎜1 + ⎟ + ⎜1 + ⎟ y = ⎜1 − ⎟ e 2 x ⎝ ⎝ x⎠ ⎝ x⎠ x⎠ where

2 ⎛ 1⎞ P = −2 ⎜1 + ⎟ , Q = 1 + , ⎝ x⎠ x

(5.12)

⎛ 2⎞ R = ⎜1 − ⎟ e 2 x ⎝ x⎠

2 2 Since 1 + P + Q = 1 − 2 − + 1 + = 0 , y = e x is a part of the C. F. x x of (5.12). Let y = v e x, where v = v ( x ) be the general solution of (5.12). Substituting y = ve x in (5.12) it reduces to d 2v ⎛ 2 du ⎞ dv R +⎜P + = , where u = e x ⎟⎠ 2 ⎝ dx u dx dx u ⇒

d 2v ⎛ 2 2 ⎞ dv ⎛ 2 ⎞ x + ⎜ −2 − + x e x ⎟ = 1− ⎟ e 2 ⎠ dx ⎜⎝ dx ⎝ x e x⎠

d 2 v 2 dv ⎛ 2 ⎞ x − = ⎜1 − ⎟ e dx 2 x dx ⎝ x⎠ 2 dv d v dp = p , so that = Let dx dx 2 dx ⇒

(5.13) (5.14)

From (5.13) and (5.14) we get dp 2 ⎛ 2⎞ − p = ⎜1 − ⎟ e x ⎝ dx x x⎠

Ch05.indd 12

(5.15)

12/14/2011 11:05:43 AM

Linear Equation of the Second Order with Variable Coefficients

5-13

which is a linear equation in p. The integrating factor is 1 1 1 log −2 dx e ∫ x = e −2log x = e x2 = 2 x 1 1 ⎛ 2⎞ Its solution is p 2 = ∫ 2 ⎜1 − ⎟ e x dx + c1 x x ⎝ x⎠

= ∫ ( x −2 e x − 2 x −3e x ) dx + c1 = x −2 e x + c1

p=

∴ ∫ e x ⎡⎣ f ( x ) + f ′ ( x )⎤⎦ dx = e x f ( x )

dv = e x + c1 x dx

Integrating ⎛ x3 ⎞ v = e x + c1 ⎜ ⎟ + c2 ⎝ 3⎠ The general solution of (5.12) is x3 y = ve x = c1 e x + c2 e x + e 2 x 3

„

d2 y dy ( x − 2) − 2 y = x 2 . 2 dx dx Solution Dividing by x, the given equation in the standard form is Example 5.1.11 Solve x

d 2 y ⎛ 2 ⎞ dy 2 + ⎜1 − ⎟ − y = x2 dx 2 ⎝ x ⎠ dx x 2 2 P =1− , Q=− , R = x2 where x x 2 2 1− P + Q = 1−1+ − = 0 Since x x y = e − x is a part of the C. F. of (5.16)

(5.16)

Let y = v e − x , where v = v ( x ) be the general solution of (5.16). Substituting y = ve − x in (5.16) it reduces to d 2v ⎛ 2 du ⎞ dv R +⎜P + = , where u = e − x ⎟ 2 dx ⎝ u dx ⎠ dx u ⇒

Ch05.indd 13

d 2v ⎛ 2 2 dv x2 −x ⎞ + − + − = 1 e ( ) ⎜ ⎟ ⎠ dx e − x2 dx 2 ⎝ x e− x

12/14/2011 11:05:44 AM

5-14

Differential Equations

d 2v ⎛ − ⎜1 + dx 2 ⎝

⇒

2 ⎞ dv = x 2e x ⎟ x ⎠ dx

dv d 2 v dp = p so that = dx dx 2 dx From (5.17) and (5.18) we get

Let

dp ⎛ 2 ⎞ − ⎜1 + ⎟ P = x 2 e x dx ⎝ x⎠

(5.17) (5.18)

(5.19)

This is a linear equation in p. 2⎞ e− x ⎛ I.F. = exp ∫ ⎜ −1 − ⎟ dx = exp( − x − 2log x ) = 2 ⎝ x⎠ x The solution of (5.19) is p

e− x e− x ⎞ 2 x ⎛ = x e dx + c1 = x + c1 ⎜ ⎝ x 2 ⎟⎠ x2 ∫ P=

∴

dv = x 3e x + c1 x 2 e x dx

Integrating v = ∫ x 3e x dx + c1 ∫ x 2 e x dx + c2

= x 3e x − 3∫ x 2 e x + c1 ∫ x 2 e x dx + c2 = x 3e x + (c1 − 3)∫ x 2 e x dx + c2

Now ∴

∫ x e dx = ( x 2 x

2

− 2 x + 2)e x

v = x 2 e x + (c1 − 3)( x 2 − 2 x + 2)e x + c2

Hence the general solution of (5.16) is y = ve − x = (c1 − 3)( x 2 − 2 x + 2) + c2 e − x + x 3

„

Example 5.1.12 Solve x 2 y ′′ − 2 x(1 + x ) y ′ + 2(1 + x ) y = x 3 . Solution Dividing by x 2 the given equation in the standard form is ⎛1 ⎞ ⎛ 1 1⎞ y ′′ − 2 ⎜ + 1⎟ y ′ + 2 ⎜ + ⎟ y = x ⎝x ⎠ ⎝ x x⎠

Ch05.indd 14

(5.20)

12/14/2011 11:05:45 AM

Linear Equation of the Second Order with Variable Coefficients

2 2 + , R= x x2 x 2 2 Since P + Qx = − − 2 + + 2 = 0 x x y = x is a part of the C. F. of (5.20). Let y = vx where v = v( x ) be the general solution of (5.20). Substituting y = vx in (5.20) it reduces to

where

P=−

2 − 2, x

5-15

Q=

d 2v ⎛ 2 du ⎞ dv R +⎜P + = , where u = x ⎟ 2 dx ⎝ u dx ⎠ dx u ⇒

d 2v ⎛ 2 + ⎜− −2+ dx 2 ⎝ x

2 ⎞ dv x = ⎟ x ⎠ dx x

d 2v dv − 2 =1 2 dx dx 2 dv d v dp = p so that = Let dx dx 2 dx dp From (5.21) and (5.22) we get − 2p =1 dx −2 dx This is a linear equation I .F . = e ∫ = e −2 x The solution of (5.23) is 1 pe −2 x = ∫ e −2 x dx + c1 = − e −2 x + c1 2 dv 1 p= = − + c1e 2 x ∴ dx 2 2x x ⎛e ⎞ Integrating v = − + c1 ⎜ +c ⎝ 2 ⎟⎠ 2 2 ⇒

(5.21) (5.22) (5.23)

∴ The general solution of (5.20) is y = vx i.e.

y = c1

xe 2 x x2 + c2 x − 2 2

„

d2 y dy − ( x 2 + 2 x ) + ( x + 2) y = x 3e x . dx 2 dx Solution The standard form of the given equation is

Example 5.1.13 Solve x 2

d2 y ⎛ − ⎜1 + dx 2 ⎝

Ch05.indd 15

2 ⎞ dy ⎛ 1 2 ⎞ + ⎜ + ⎟ y = xe x ⎟ x ⎠ dx ⎝ x x 2 ⎠

(5.24)

12/14/2011 11:05:45 AM

5-16

Differential Equations

⎛ Here P + Qx = − ⎜1 + ⎝

2⎞ ⎛ 1 2 ⎞ ⎟ +⎜ + ⎟ x=0 x ⎠ ⎝ x x2 ⎠

∴ y = x is a part of the C.F. of (5.24) Putting y = vx,

dy dv = x + v, dx dx

d2 y d 2v dv = x +2 in (5.24) 2 2 dx dx dx

we get d 2 v dv dp dv − = e x or − P = e x , where P = 2 dx dx dx dx This is linear in p. − dx I.F. = e ∫ = e − x pe − x = ∫ e x ⋅ e − x dx + c1 = x + c1

∴

⇒

P=

dv = xe x + c1e x dx

Integrating we get v = xe x − e x + c1e x + c2 ∴ The complete solution of (5.24) is y = vx = x 2 e x − xe x + c1 xe x + c2 x

„

EXERCISE 5.1 Solve the following differential equations: 1.

d2 y dy − x2 + xy = x. 2 dx dx Ans: y = c1 x + c2 x ∫ x −2 e 3 x dx + 1 1 3

d2 y dy − (3 + x ) + 3 y = 0. 2 dx dx Ans: y = c1e x + c2 ( x 3 + 3 x 2 + 6 x +  )

2. x

d2 y dy + (1 − x ) = y + e x . 2 dx dx Ans: y = e x log x + c1e x ∫ x −1e − x dx + c3e x

3. x

Ch05.indd 16

12/14/2011 11:05:46 AM

Linear Equation of the Second Order with Variable Coefficients

5-17

d2 y dy − 2( x + 3) + ( x + 5) y = e x . 2 dx dx 1 x 5 Ans: y = c1e ( x + 1) + c2 e x − xe x 2 d2 y dy 5. (3 − x ) 2 − (9 − 4 x ) + (6 − 3 x ) y = 0. dx dx Ans: y = c1e x + c2 e 3 x (4 x 3 − 42 x 2 + 150 x − 183)

4. ( x + 1)

6. x 2

d3 y d2 y dy 2 + x − 2 x + 2 y = 0. 3 2 dx dx dx

Ans: y = c1 x 2 + c2 x + c3 ( x 2 ∫ x −3e − x dx − x ∫ x −2 e − x dx ) d2 y dy + (1 − x ) − y = e x . 2 dx dx e− x Ans: y = e x log x + c1e x ∫ dx + c2 e x x

7. x

dy ⎛ d2 y ⎞ − y = ( x − 1) ⎜ 2 − x + 1⎟ . ⎝ ⎠ dx dx Ans: y = − x 2 − 1 + c1e x + c2 x

8. x

d2 y dy + x − y = x(1 − x 2 )3 2 . 2 dx dx 1 Ans: y = − x(1 − x 2 )3 2 − c1{(1 − x 2 )1 2 + x sin −1 x} + c2 x 9 d2 y dy 10. ( x sin x + cos x ) 2 − x cos x + y cos x = sin x( x sin x + cos x )2 . dx dx 1 1 Ans: y = x cos 2 x − sin 2 x − c1 cos x + c2 x 4 2 d2 y dy 11. x( x cos x − 2sin x ) 2 + ( x 2 + 2)sin x − 2( x sin x + cos x ) y = 0. dx dx 2 Ans: y = c1 sin x + c2 x 9. (1 − x 2 )

12. (2 x − 1)

d2 y dy − 2 + (3 − 2 x ) y = 2e x . 2 dx dx

Ans: y = − c1 xe − x − xe x + e x ∫ (2 x − 1) log(2 x − 1)dx −2e x ∫ [e −2 x (2 x − 1)∫ e 2 x log(2 x − 1)dx ]dx + c2 e x

Ch05.indd 17

12/14/2011 11:05:46 AM

5-18

Differential Equations

d 2y dy +P +Qy = R 2 dx dx BY CHANGING THE DEPENDENT VARIABLE AND REMOVING THE FIRST DERIVATIVE REDUCTION TO NORMAL FORM

5.2

GENERAL SOLUTION OF

Let y = uv

d2 y dy + P + Qy = R 2 dx dx dy dv du ⇒ =v +u dx dx dx 2 2 d y d v du dv d 2u = u + 2 + v dx 2 dx 2 dx dx dx 2

(5.25) (5.26) (5.27)

Substituting in (5.25) rearranging the terms d 2v ⎛ du ⎞ dv ⎛ d 2 u du ⎞ + Pu + 2 +⎜ 2 +P + Qu⎟ v = R ⎜ ⎟⎠ 2 ⎝ ⎝ ⎠ dx dx dx dx dx dv The term will disappear if we choose such that dx du − 1 pdx Pu + 2 x = e 2∫ = 0 i.e. dx u

(5.28)

(5.29)

d 2u 1 du 1 dp 1 2 1 dp =− P − u = P u− u 2 2 dx 2 dx 4 2 dx dx (5.30) Substituting (5.29) and (5.30) into (5.28) we get du 1 = − Pu 2 dx

u

⇒

d 2v ⎛ 1 2 1 dp 1 2 ⎞ +⎜ P u− u − P u + Qu⎟ v = R 2 ⎠ dx ⎝ 4 2 dx 2 ⇒

d 2v ⎛ 1 1 dp ⎞ R + ⎜ Q − P2 − ⎟v= dx 2 ⎝ 4 2 dx ⎠ u

⇒ where I =Q−

d 2v R + Iv = = 5 2 dx u

(5.31)

1 2 1 dP P − , is called the normal form of (5.25). 4 2 dx

After obtaining v from (5.31) the general solution of (5.25) is y = uv − 1 Pdx where u = e 2 ∫ .

Ch05.indd 18

12/14/2011 11:05:47 AM

Linear Equation of the Second Order with Variable Coefficients

Example 5.2.1 Solve

5-19

d2 y dy 2 − 4 x + 4 x2 y = ex . 2 dx dx

d2 y dy 2 − 4 x + 4 x2 y = ex . 2 dx dx Comparing with the standard equation

Solution

P = −4 x,

Q = 4 x2 ,

(5.32) R = ex

2

− 1 Pdx 2 xdx 2 = ex Let u = e 2 ∫ = e ∫ If y = uv, Eq. (5.32) reduces to the normal form

d 2v R + Iv = 2 dx u

(5.33)

where I =Q−

1 2 1 dp ⎛ 1⎞ p − = 4 x 2 − 4 x 2 + ⎜ − ⎟ ( −4) = 2 ⎝ 2⎠ 4 2 dx 2

R ex = 2 =1 u ex Substituting these value in (5.33) we get d 2v + 2v = 1 dx 2 A.E. is m2 + 2 = 0 ⇒ m = ± 2i vc = c1 cos x 2 + c2 sin x 2 1 1 vP = 2 ⋅1 = D +2 2 S=

(5.34)

The general solution of (5.34) is v = c1 cos x 2 + c2 sin x 2 +

1 2

Hence the general solution of (5.32) is 1⎤ 2 ⎡ y = uv = e x ⎢c1 cos x 2 + c2 sin x 2 + ⎥ 2⎦ ⎣

d 2 y 2 dy ⎛ 2⎞ − + ⎜1 + 2 ⎟ y = xe x . dx 2 x dx ⎝ x ⎠ d 2 y 2 dy ⎛ 2⎞ − + ⎜1 + 2 ⎟ y = xe x 2 dx x dx ⎝ x ⎠

„

Example 5.2.2 Solve Solution

Ch05.indd 19

(5.35)

12/14/2011 11:05:47 AM

5-20

Differential Equations

Comparing (5.35) with the standard equation we have 2 2 P=− , Q = 1 + 2 , R = xe x x x To remove the first derivative choose − 1 pdx − 1 −2 dx u = e 2 ∫ = e 2 ∫ x = e log x = x

If we take y = uv, Eq. (5.35) reduces to d 2v + Iv = S dx 2

(5.36)

Where 1 2 1 dp 2 1⎛ 4 ⎞ 1⎛ 2 ⎞ P − = 1+ 2 − ⎜ 2 ⎟ − ⎜ 2 ⎟ = 1 4 2 dx x 4⎝ x ⎠ 2⎝ x ⎠ R xe x S= = = ex u x Eq. (5.36) now becomes ( D 2 + 1) v = e x I =Q−

(5.37)

vc = c1 cos x + c2 sin x ⎫ ex ⎪ cos sin v = c x + c x + 1 1 ⎬ 1 2 2 vp = 2 ex = ex ⎪ 2 ⎭ D +1 ∴

ex ⎤ ⎡ y = uv = x ⎢(c1 cos x + c2 sin x ) + ⎥ . 2⎦ ⎣

„

d2 y dy + 4 x5 + ( x 8 + 6 x 4 + 4) = 0. 2 dx dx Solution The given equation, in the standard form, is

Example 5.2.3 Solve 4 x 2

d2 y dy 1 ⎛ 6 4⎞ + x3 + ⎜ x + 6x2 + 2 ⎟ y = 0 2 ⎝ dx dx 4 x ⎠

(5.38)

1⎛ 6 4⎞ 2 ⎜⎝ x + 6 x + 2 ⎟⎠ , R = 0 4 x To remove the first derivative choose

Here P = x 3 , Q =

u=e ∫

− 12 pdx

=x

− x4 2

If we take y = uv Eq. (5.38) reduces to the normal from d 2v + Iv = S dx 2

Ch05.indd 20

(5.39)

12/14/2011 11:05:48 AM

Linear Equation of the Second Order with Variable Coefficients

5-21

where I =Q−

1 2 1 dp 1 ⎛ 6 4⎞ 1 1 1 P − = ⎜ x + 6 x 2 + 2 ⎟ − x 6 − 3x 2 = 2 4 2 dx 4 ⎝ x ⎠ 4 2 x

R =0 u Substituting these values in (5.39) we get d 2v 1 d 2v 2 + v = 0 ⇒ x +v=0 dx 2 x 2 dx 2 S=

Put x = e z so that z = log x x

d = θ, dx

x2

d2 d⎞ ⎛ = θ (θ − 1) ⎜ θ = 2 ⎟ ⎝ dx 2 d ⎠

we obtain

θ (θ − 1)v + v = 0 ⇒ (θ 2 − θ + 1)v = 0 A.E. is m2 − m + 1 = 0

⇒

m=

1± i 3 2

The general solution is z ⎛ ⎛ 3⎞ ⎛ 3⎞⎞ v = e 2 ⎜ c1 cos ⎜ z + c2 sin ⎜ z ⎟ ⎟⎟ ⎝ 2 ⎠ ⎝ 2 ⎠⎠ ⎝ 1 ⎛ ⎛ 3 ⎞ ⎛ 3 ⎞⎞ log x ⎟ + c2 sin ⎜ log x ⎟ ⎟ = x 2 ⎜ c1 cos ⎜ ⎝ 2 ⎠ ⎝ 2 ⎠⎠ ⎝

Hence the complete solution of the differential equation is y=e

− x4 8

⎡ ⎛ 3 ⎞ ⎛ 3 ⎞⎤ x ⎢c1 cos ⎜ log x ⎟ + c2 sin ⎜ log x ⎟ ⎥ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎢⎣

„

d2 y dy − 2( x 2 + x ) + ( x 2 + 2 x + 2) y = 0. 2 dx dx Solution The given equation may be written in the form

Example 5.2.4 Solve x 2

d2 y ⎛ 1 ⎞ dy ⎛ 2 2 ⎞ − 2 ⎜1 + ⎟ + ⎜1 + + 2 ⎟ y = 0 2 ⎝ dx x ⎠ dx ⎝ x x ⎠ 2 2 ⎛ 1⎞ Here P = −2 ⎜1 + ⎟ , Q = 1 + + 2 ⎝ x⎠ x x − 1 pdx (1+ 1 )dx Hence u = e 2 ∫ = e ∫ x = e x + log x = xe x

Ch05.indd 21

12/14/2011 11:05:48 AM

5-22

Differential Equations

1 dp 1 2 − p =0 2 dx 4 The transformed equation is d 2v dv =0 ⇒ =c or v = c1 x + c2 2 dx dx 1

∴

I =Q−

The general solution of the given equation is y = uv = xe x (c1 x + c2 ) = e x (c1 x 2 + c2 x ). Example 5.2.5 Solve

„

d2 y 1 dy ⎛ 1 1 6⎞ + 13 + ⎜ 2 3 − 4 3 − 2 ⎟ y = 0. 2 dx x dx ⎝ 4 x 6x x ⎠

Solution Here P = x −1 3 ,

Q=

1 pdx 2

3

1 1 6 − 43 − 2 23 4x 6x x

2

− x3 Hence u=e ∫ =e 4 1 1 6 1 1 1 1 6 ∴ I = 23 − 43 − 2 + − =− 2 43 23 4x 6x x 6x 4x x The transformed equation is d 2 v 6v d 2v 2 − = 0 or x − 6v = 0 dx 2 x 2 dx 2 This is a homogeneous linear equation and its solution is v = c1 x 3 + c2 x −2 −

The general solution of the given equation is 2

y = uv = e − 4 x 3 (4 x 3 + c2 x −2 ) 3

„

d ⎛ 2 dy ⎞ 2 ⎜ cos x ⎟⎠ + y cos x = 0. dx ⎝ dx Solution On simplification the given equation becomes d2 y dy − 2 tan x + y = 0 2 dx dx Here P = −2 tan x, Q =1

Example 5.2.6 Solve

∴

Ch05.indd 22

1

− pdx u = e 2 ∫ = sec x 1 dp 1 2 I =Q− − p =2 2 dx 4

12/14/2011 11:05:49 AM

Linear Equation of the Second Order with Variable Coefficients

5-23

d 2v + 2v = 0 dx 2 whose solution is v = c1 cos x 2 + c2 sin x 2 The reduced equation is

∴ y = uv = sec x(c1 cos x 2 + c2 sin x 2) Example 5.2.7 Solve

„

d2 y dy 2 − 4 x + (4 x 2 − 1) y = −3e x sin 2 x. 2 dx dx

Solution Here P = −4 x,

Q = 4 x2 − 1 1

pdx − 2 xdx 2 u = e 2 ∫ = e∫ = ex

∴

1 dp 1 2 1 1 − p = 4 x 2 − 1 − ( −4) − ⋅ 16 x 2 = 1 2 dx 4 2 4 Hence the transformed equation is I =Q−

1 d 2v −4 xdx x2 2∫ + = − = −3sin 2 x 3 sin 2 . v e x e 2 dx Here the complementary function is c1 cos x + c2 sin x

P.I. =

1 −3 x sin 2 x ( −3sin 2 x ) = = sin 2 x D +1 −4 + 1 2

The solution of this equation is v = c1 cos x + c2 sin x + sin 2 x The complete primitive of the given equation is y = uv = e x (4 cos x + c2 sin x + sin 2 x ) 2

„

EXERCISE 5.2 Solve: 1.

d 2 y 2 dy ⎛ 2⎞ − + ⎜1 + 2 ⎟ y = xe x . 2 dx x dx ⎝ x ⎠ 1 ⎞ ⎛ Ans: y = x ⎜ c1 cos x + c2 sin x + e x ⎟ ⎝ 2 ⎠

Ch05.indd 23

12/14/2011 11:05:49 AM

5-24

2.

Differential Equations

d2 y dy ⎛ 6⎞ + 4 cosec 2 x + ⎜ 2sec2 x − 2 ⎟ y = 0. 2 ⎝ dx dx x ⎠ Ans: y = (c1 x −3 + c2 x −2 ) cot x

3.

2 d2 y dy 2 −x 2 + 2 + ( + 5) = . x x y xe dx 2 dx

1 ⎞ 2 ⎛ Ans: y = ⎜ c1 cos 2 x + c2 sin 2 x + x ⎟ e − x 2 ⎝ 4 ⎠ 4.

d2 y 1 dy 1 − + 2 ( x + x − 8) y = 0. 2 dx x dx 4 x 1⎞ ⎛ Ans: y = x ⎜ c1 x 2 + c2 ⎟ ⎝ x⎠

5.

d2 y dy − 2 tan x + 5 y = 0. 2 dx dx Ans: y = sec x[c1 cos 6 x + c2 sin 6 x ]

6. x 2

d2 y dy − 2( x 3 + x ) + ( x 2 + 2 x + 2) y = 0. 2 dx dx

Ans: y = xe x (c1 x + c2 ) 7.

d2 y dy + 4 x + 4 x 2 y = 0. 2 dx dx Ans: y = e − x (c1e x 3

8.

2

+ c2 e − x 2 )

d 2 y 2 dy + + x 2 y = 0. dx 2 x dx Ans: y = (c1 sin nx + c2 cos nx ) x

9.

10.

d 2 y 2 dy + = n 2 y. dx 2 x dx Ans: xy = c1e nx + c2 e − nx d2 y dy − 2bx + b 2 x 2 y = 0. 2 dx dx 1

x2

Ans: y = ce 2 sin( x b + A)

Ch05.indd 24

12/14/2011 11:05:50 AM

Linear Equation of the Second Order with Variable Coefficients

5-25

d2 y dy + 4 x5 + ( x 8 + 6 x 4 + 4) y = 0. dx 2 dx − x4 ⎛ 3 ⎞ Ans: y = ce 8 x sin ⎜ log x + 4⎟ ⎝ 2 ⎠ 2 d y dy 12. − 2 tan x + 5 y = sec xe x . 2 dx dx

11. 4 x 2

1 Ans: y = (c1 sin 6 x + c2 cos 6 x )sec x + e x sec x 7

5.3

d 2y d y +Qy = R P dx 2 dx BY CHANGING THE INDEPENDENT VARIABLE

GENERAL SOLUTION OF

The differential equation is d2 y dy + P + Qy = R 2 dx dx

(5.40)

Let the independent variable x be changed to z where z = z ( x ). We have 2

dy dy dz = dx dz dx

d 2 y d 2 y ⎛ dz ⎞ dy d 2 z = + ⎜ ⎟ dx 2 dz 2 ⎝ dx ⎠ dz dx 2

and

Substituting in (5.40) 2

dy dz ⎛ dz ⎞ d 2 y dy d 2 z + +P + Qy = R ⎜⎝ ⎟⎠ 2 2 dx dx dx dx dz dx 2

dz ⎞ dy ⎛ dz ⎞ d 2 y ⎛ d 2 z +⎜ 2 +P ⎟ + Qy = R ⎜⎝ ⎟⎠ 2 ⎝ dx dz dx dx ⎠ dz 2

dz Dividing by ⎛⎜ ⎞⎟ we can write it as ⎝ dx ⎠ d2 y dy + P1 + Q1 y = R1 2 dz dx where

( P=

d2z dx 2

1

Q1 =

Ch05.indd 25

dz + P dx )

( dxdz )2

(5.41)

(5.42)

Q

( dxdz )2

(5.43)

12/14/2011 11:05:51 AM

5-26

Differential Equations

R1 =

R (5.44)

( )

dz 2 dx

Here P1, Q1, R1 are functions of x, which can be changed into functions of z using z = z ( x ). Then we have to choose z so that P1 = 0 or Q = a 2 (constant). Case 1 If P1 = 0 we have d2z dx 2 dz dx

d2z dz +P =0 ⇒ 2 dx dx

( )

= −P

Integrating we get ⎛ dz ⎞ log ⎜ ⎟ = − ∫ pdx ⎝ dx ⎠

dz − pdx =e ∫ dx

⇒

Integrating again − pdx z = ∫ e ∫ dx

Case 2 If Q1 = a 2 then from (5.43) we have 2

dz ⎛ d2 ⎞ Q = a2 ⎜ ⎟ ⇒ = ± aQ2 ⇒ z = ∫ ± aQ2 dx ⎝ dx ⎠ dx Here we have to take ± sign to make

( dxdz )

2

real.

Usually we can take a 2 = 1 or take a2 such that a2 = 1. Thus we get the value of z. In this case P1 should either be zero or constant. Then only Eq. (5.41) can be solved. Q

d 2 y dy − + 4 x2 y = x5 . dx 2 dx Solution Putting the equation in the standard form

Example 5.3.1 Solve x

d 2 y 1 dy − + 4 x2 y = x4 dx 2 x dx Here

Ch05.indd 26

(5.45)

1 P = − , Q = 4 x2 , R = x4 x

12/14/2011 11:05:51 AM

Linear Equation of the Second Order with Variable Coefficients

5-27

choose z such that 2

⎛ dz ⎞ 2 ⎜⎝ ⎟⎠ = Q = 4 x dx

⇒

dz = 2x ⇒ z = x2 dx

(5.46)

Eq. (5.45) is transformed into d2 y dy + P1 + Q1 y = R1 2 dz dz where

( P=

d2z dx 2

1

Q1 = and

R1 =

R

( dxdz )2

dz + P dx )

( dxdz )2 Q

( dxdz )2 =

=

=

(5.47)

2 − 1x ⋅ 2 x =0 4 x2

4 x2 =1 4 x2

x4 x2 z = = 4 x2 4 4

Substituting these values in (5.47) we get d2 y 1 +y= z 2 dz 4

or

(θ 2 + 1) =

1 z 4

1⎞ ⎛ ⎜⎝ θ ≡ ⎟⎠ dz

(5.48)

yc = c1 cos z + c2 sin z 1 1 1 z yp = z = (1 − θ 2 −  ) z = 2 4 θ +1 4 4 The complete solution of Eq. (5.48) is y = c1 cos x 2 + c2 sin x 2 +

1 2 x 4

„

d 2 y dy − − 4 x 3 y = 8 x 3 sin x 2 . dx 2 dx Solution Putting the equation in the standard form

Example 5.3.2 Solve x

d 2 y 1 dy − − 4 x 2 y = 8 x 2 sin x 2 dx 2 x dx 1 P = − , Q = −4 x 2 , R = 8 x 2 sin x 2 Here x 2 ⎛ dz ⎞ choose z such that ⎜ ⎟ = Q = −4 x 2 ⎝ dx ⎠

Ch05.indd 27

(5.49)

12/14/2011 11:05:52 AM

5-28

Differential Equations

dz = 2x ⇒ z = x2 dx Eq. (5.49) now transforms into Discarding the – sign, we get

d2 y dy + P1 + Q1 y = R1 2 dz dz where

( P=

d2z dx 2

( )

dz 2 dx

1

Q1 = and

R1 =

dz + P dx )

Q

( )

dz 2 dx

R

( )

dz 2 dx

=

=

=

2 + ( − 1x ) ⋅ 2 x 4 x2

(5.50)

=0

−4 x 2 = −1 4 x2

8 x 2 sin x 2 = 2sin x 2 = 2sin z 4 x2

Substituting these values in (5.50) d2 y − y = 2sin z dz 2 (θ 2 − 1) y = 2sin z

d⎞ ⎛ ⎜⎝ θ ≡ ⎟⎠ dz

yc = c1e z + c2 e − z 1 1 2sin z = 2 2 sin z = − sin z yp = 2 θ −1 −1 − 1 The complete solution of (2.49) is 2 2 y = c1e x + c2 e − x − sin x 2

„

d2 y dy x − (1 + 4e x ) + 3e 2 x y = e 2( x + e ) . 2 dx dx 2 d y dy x − (1 + 4e x ) + 3e 2 x y = e 2( x + e ) Solution 2 dx dx

Example 5.3.3 Solve

Here

P = −(1 + 4e x ), Q = 3e 2 x , R = e 2( x + e

choose z such that

x

(5.51)

)

dz = ± aQ2 = 3e 23x = e x dx

(5.52)

Integrating z = e x (5.52). Now Eq. (5.51) reduces to d2 y dy + P1 + Q1 y = R1 2 dx dx

Ch05.indd 28

(5.53)

12/14/2011 11:05:52 AM

Linear Equation of the Second Order with Variable Coefficients

5-29

where P1 = Q1 = R1 =

d2z dx 2

dz + P dx

( )

dz 2 dx

=

e x − (1 + 4e x )e x = −4 e2 x

2x

Q

( )

= 3e e2 x = 3

dz 2 dx

e2 x ⋅ e2e x = = e2e = e2 z 2x e x

E

( dxdz )2

Substituting these values in (5.53) we get d2 y dy − 4 + 3 y = e2 z 2 dz dz (θ 2 − 4θ + 3) y = e 2 z

or

yc = c1e z + c2 e 3 z e2 z 1 yp = 2 e2 z = 2 = −e2 z 2 − 4⋅2 + 3 θ − 4θ + 3

(5.54)

The complete solution of (5.54) is y = c1e z + c2 e 3 z − e 2 z Hence the complete solution of (5.51) is y = c1 exp (e x ) + c2 exp (3e x ) − exp (2e x ) Example 5.3.4 Solve

„

d2 y dy − (8e 2 x + 2) + 4e 4 x y = e6 x . 2 dx dx

dz = 2e 2 x dx d2z i.e. z = e 2 x and 2 = 4e 2 x the equation becomes dx d2 y dy 1 −4 + y= z 2 dz dz 4

Solution Putting

whose C.F. is c1e (2+ P.I. =

Ch05.indd 29

3)z

+ c2 e (2 −

3)z

1 1 1 z = ( z + 4) 2 4 1 − 4D + D 4

12/14/2011 11:05:52 AM

5-30

Differential Equations

Hence the solution is y = c1 e (2+

+ c2 e (2 −

3 ) e2 x

3 ) e2 x

1 + e2 x + 1 4

„

d2 y dy + cot x + 4 y cosec2 x = 0. 2 dx dx 2 ⎛ dz ⎞ Solution Let us choose z such that ⎜ ⎟ = 4 cosec2 x. ⎝ dx ⎠ This implies that dz = 2cosec x dx or z = 2log tan 2x

Example 5.3.5 Solve

∴

P1 =

dz P ( dx )+

( dxdz )2

d2z dx 2

=0

∴ Changing the independent variable from x to z the equation becomes d2 y + y=0 dz 2 whose solution is y = A cos( z + B) ∴ The solution of the give equation is y = A cos(2log tan 2x + B)

„

d2 y dy 1 + 3x 5 + a2 y = 2 . 2 dx dx x Solution The equation may be written in the form

Example 5.3.6 Solve x 6

d 2 y 3 dy a 2 1 + + 6 y= 8 2 dx x dx x x 2

a2 a ⎛ dz ⎞ Let us choose z such that ⎜ ⎟ = 6 or z = − 2 ⎝ dx ⎠ x 2x Here R1 =

R

( dxdz )2

=

P1 =

Ch05.indd 30

1 x6 1 2z = 2 2 =− 3 2 8a x a a dz P dx +

( dxdz )2

d2z dx 2

=

3 x

⋅ xa3 + ( − 3xa4 )

( ) 2 x3

2

=0

12/14/2011 11:05:53 AM

Linear Equation of the Second Order with Variable Coefficients

5-31

The transformed equation is d2 y 2z +y=− 3 2 dz a The C.F. is c1 cos z + c2 sin z = c1 cos P.I. =

( ) + c sin ( ) a 2 x2

a 2 x2

2

1 ⎛ 2 ⎞ 2 2 1 2 ⎜⎝ − 3 z ⎟⎠ = − 3 (1 − D  ) z = − 3 z = 2 2 D +1 a a a a x 2

y = c1 cos

∴

( ) + c sin ( ) + a 1x a 2 x2

a 2 x2

2

2

„

2

d 2 y dy − − 4 x 2 y = 8 x 3 sin x 2 . dx 2 dx Solution The equation may be written in the form

Example 5.3.7 Solve x

d 2 y 1 dy − − 4 xy = 8 x 2 sin x 2 dx 2 x dx 2

⎛ dz ⎞ Find z such that ⎜ ⎟ = 4 x 2 or z = x 2 ⎝ dx ⎠ Here R1 =

R

( )

dz 2 dx

=

8 x 2 sin x 2 = 2sin x 2 = 2sin z and Q1 = 0 4 x2

The transformed equation is d2 y − y = 2sin z dz 2 whose C.F. = c1e z + c2 e − z = c1e x + c2 e − x 2

P.I. =

2

1 2sin z = − sin z = − sin x 2 D −1 2

y = c1e x + c2 e − x − sin x 2

∴

2

2

„

d2 y dy cos x + sin x − 2 y cos 2 x = 2cos5 x. 2 dx dx Solution Dividing by cos x we get

Example 5.3.8 Solve

d2 y dy + tan x − 2 y cos 2 x = 2cos 4 x 2 dx dx

Ch05.indd 31

12/14/2011 11:05:53 AM

5-32

Differential Equations 2

⎛ dz ⎞ Find z such that ⎜ ⎟ = cos 2 x or z = sin x ⎝ dx ⎠ Here R1 =

R

( dxdz )2

=

2cos 4 x = 2cos 2 x = (1 − z 2 ) cos 2 x

The transformed equation is d2 y − 2 y = 2(1 − z 2 ) dx 2 whose C.F. is c1e P.I. =

2z

+ c2 e −

2z

= c1e

2 sin x

+ c2 e −

2 sin x

1 1⎛ 1 ⎞ 2(1 − z 2 ) = − ⎜1 + D 2  ⎟ 2(1 − z 2 ) = z 2 = sin 2 x ⎝ ⎠ D −1 2 4 2

The complete solution is y = c1e

2 sin x

+ c2 e −

2 sin x

+ sin 2 x

„

EXERCISE 5.3 Solve: 1.

d2 y dy − cot x − sin 2 x y = 0. 2 dx dx Ans. y = c1 exp ( − cos x ) + c2 exp (cos x )

2.

d2 y dy + tan x + cos 2 x y = 0. 2 dx dx Ans. y = c1 cos(sin x ) + c2 sin (sin x )

d 2 y 2 dy a 2 + + y = 0. dx 2 x dx x 4 a a Ans. y = c1 cos + c2 sin x x 2 d y dy − (1 + 4e x ) + 3e 2 x y = 0. 4. 2 dx dx

3.

Ans. y = c1 exp (e x ) + c2 exp (3e x )

Ch05.indd 32

12/14/2011 11:05:54 AM

Linear Equation of the Second Order with Variable Coefficients

5. (1 + x 2 )

d2 y dy + x + 2 y = 0. dx 2 dx

Ans. y = c1 cos 6.

(

2 sinh −1 x + c2

5-33

)

d2 y dy − cot x − sin 2 xy = cos x − cos 2 x. dx 2 dx Ans. y = c1 exp ( − cos x ) + c2 exp (cos x ) − cos x

7. (1 − x 2 )

d2 y dy − x + m2 y = 0. 2 dx dx

Ans. y = c1 sin ( n x 2 − 1 + α ) d2 y dy + sin x cos x + 4 y = 0. 2 dx dx x ⎛ ⎞ Ans. y = c cos ⎜ log tan + A⎟ ⎝ ⎠ 2

8. sin 2 x

9. (1 + x 2 )2

d2 y dy + 2 x (1 + x 2 ) + 4 y = 0. dx 2 dx

Ans. y(1 + x 2 ) = c1 (1 − x 2 ) + c2 x

Ch05.indd 33

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Ch05.indd 34

12/14/2011 11:05:55 AM

6 Integration in Series: Legendre, Bessel and Chebyshev Functions 6.1 LEGENDRE FUNCTIONS 6.1.1 Introduction If a homogeneous linear differential equation of the form L( y ) = ( ∑ ar D r ) y = 0 has constant coefficients ar , it can be solved by elementary methods, and the functions involved in solutions are elementary functions such as x n , e x ,log x,sin x etc. However, if such an equation has variable coefficients and functions of the independent variable x, it has to be solved by other methods. Legendre’s equation and Bessel’s equation are two very important equations of this type. We take up below, solution of these equations by application of two standard methods of solution: 1. The power series method and 2. The Frobenius1 method, which is an extension of the power series method.

6.1.2

Power Series Method of Solution of Linear Differential Equations

The power series method, which gives solutions in the form of power series, is the standard method for solving ordinary linear differential equations with variable coefficients. 1

Georg Frobenius (1849–1917), German mathematician, also made important contributions to the theory of matrices and groups.

Ch06.indd 1

12/13/2011 11:11:43 AM

6-2

Differential Equations

Power series ∞

An infinite series of the form

∑a

m= 0

m

( x − x0 ) m = a0 + a1 ( x − x0 ) +

a2 ( x − x0 )2 + where a0 , a1 , a2 , are real constants is called a power series. These constants are called the coefficients of the series. x0 is a constant and is called the centre of the series, and x is a real variable. For the choice x0 = 0, we obtain the power series ∞

∑a

m= 0

m

x m = a0 + a1 x + a2 x 2 +

(6.1)

with the origin as its centre. Convergence interval If there exists a positive real number R such that for all x in I = ( x − R, x + R) the series Eq. (6.1) is convergent, then I is called the interval of convergence for the series Eq. (6.1) and R is called the −1 a 1 radius of convergence. R is defined by R = lim ann or R = lim n . n→∞ a n→∞ n +1 Real analytic functions A real function f ( x ) is called analytic at x = x0 if it can be represented by a power series in powers of ( x − x0 ) with radius of convergence R > 0.

(

)

6.1.3 Existence of Series Solutions: Method of Frobenius The condition for the existence of power series solutions of the type ∞

∑ a (x − x ) n= 0

n

n

0

(6.2)

for a second order differential equation with variable coefficients, written in the standard form y ′′ + p( x ) y ′ + q( x ) y = 0

(6.3)

is that the coefficient functions p( x ) and q( x ) must be analytic at x = x0 (equivalently expansible in Taylor’s series or differentiable any number of times). In the case of Bessel’s equation y ′′ +

Ch06.indd 2

1 ⎛ x 2 − p2 ⎞ y′ + ⎜ y=0 ⎝ x 2 ⎟⎠ x

(6.4)

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Integration in Series

6-3

the above condition is not satisfied since 1x and x x−2p are not analytic at x = 0 and the power series method fails. The method of Frobenius which is an extension of the power series method applies to an important class of equations of the form 2

2

p( x ) q( x ) (6.5) y′ + 2 y = 0 x x where the functions p( x ) and q( x ) are analytic at x = 0, or equivalently, p( x ) and q( x ) are expansible as power series. It gives at least one solution that can be represented as y ′′ +

∞

y( x ) = x m ∑ ar x r = x m ( a0 + a1 x + a2 x 2 +

)

(6.6)

r=0

where the exponent m may be any real or complex number and m is chosen so that a0 ≠ 0. Eq. (6.5) also has a second solution, which is linearly independent of the above solution in the form Eq. (6.6) with a different m and different coefficients or a logarithmic term. Bessel’s equation (6.4) is of the form Eq. (6.5) with p( x ) = 1 and q( x ) = x 2 − n2 analytic at x = 0 so that the method of Frobenius applies. Regular and singular points The point x0 at which the coefficients p and q are analytic is called a regular point of Eq. (6.3). Then the power series method can be applied for its solution. In the case of Legendre’s equation (1 − x 2 ) y ′′ − 2 xy ′ + n( n + 1) y = 0 which can be put in the standard form y ′′ −

2x n( n + 1) y′ + y = 0 ( x ≠ 1) 2 1− x 1 − x2

the functions p( x ) = − 1−2 xx2 and q( x ) = n1(−nx+21) are expressible as a power series about x = 0 and so they are analytic at x = 0. Hence the equation can be solved by the power series method. If x0 is not regular, it is called a singular point. In respect of Bessel’s Eq. (6.4) the point x = 0 is not regular and is therefore a

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6-4

Differential Equations

singular point of Eq. (6.4). But xp( x ) = 1 and x 2 q( x ) = x 2 − p 2 are analytic at x = 0 and hence we can apply Frobenius method for the solution of Eq. (6.4). The series solution of p( x ) y ′′ + q( x ) y ′ + r ( x ) y = 0 by Frobenius method consists of the following steps: 1. Assume that y given by Eq. (6.6) is a solution of the equation. 2. Compute y′ and y′′ and substitute for y, y′ and y′′ in the equation. 3. Equate to zero the coefficient of the lowest power of x; it gives a quadratic in m, which is known as the indicial equation, giving two values for m. 4. Equate to zero the coefficients of the other powers of x and find the values of a1 , a2 , a3 , in terms of a0.

6.1.4

Legendre Functions

The power series method of solution can be applied to find the solution of the Legendre’s2 differential equation (1 − x 2 ) y ′′ − 2 xy ′ + n( n + 1) y = 0 [(1 − x 2 ) y ′ ]′ + n( n + 1) y = 0

or

(6.7)

which arises in boundary value problems having spherical symmetry. Here n is a real constant. We can put Eq. (6.7) in the standard form 2x n( n + 1) (6.8) y′ + y=0 2 1− x 1 − x2 by dividing it by the coefficient (1 − x 2 ) of y′′ in Eq. (6.7). Now, the coefficients of y′ and y in Eq. (6.8) are y ′′ −

−

2x 1 − x2

and

n( n + 1) 1 − x2

respectively, and these functions are analytic at x = 0 (i.e., derivable any number of times). Hence we may apply the power series method for its solution. 2

Adrien Marie Legendre (1752–1833), French mathematician, who became a professor in Paris in 1775 made important contributions to special functions, elliptic integrals, number theory and the calculus of variations. His book Elements de Geometrie (1794) became very famous and had 12 editions in less than 30 years.

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Integration in Series

6-5

Let us assume a power series solution of Eq. (6.7) in the form ∞

y( x ) = ∑ ar x r

(6.9)

r=0

Differentiating Eq. (6.8) w.r.t. ‘x’ twice, we have ∞

y ′( x ) = ∑ rar x r −1 , r =1 ∞

y ′′( x ) = ∑ r ( r − 1)ar x r − 2

(6.10)

r=2

Substituting the expressions for y, y′ and y′′ from Eqs. (6.9) and (6.10) in Eq. (6.7) ∞

∞

∞

(1 − x 2 )∑ r ( r − 1)ar x r − 2 − 2 x ∑ rar x r −1 + n( n + 1)∑ ar x r = 0 ∞

r=2

∞

r =1 ∞

r=0 ∞

⇒ ∑ r ( r − 1)ar x r − 2 − ∑ r ( r − 1)ar x r − 2∑ rar x r + n( n + 1)∑ ar x r = 0 r=2

r=2

r =1

r=0

(6.11)

By writing each series in the expanded form and arranging each power of x in a column, we have 2 ⋅ 1 a2 + 3 ⋅ 2 a3 x + 4 ⋅ 3 a4 x 2 + + ( r + 2)( r + 1) ar + 2 x r + − 2 ⋅ 1a2 x 2 − − r ( r − 1)ar ⋅ x r − − 2 ⋅ 1a1 x −2 ⋅ 2a2 x 2 − − 2r ⋅ ar ⋅ x r − n ( n + 1)a0 + n( n + 1)a1 x + n ( n + 1)ar x 2 + + n ( n + 1)ar ⋅ x r + = 0 Since Eq. (6.9) is a solution of Eq. (6.7), this must be an identity in x. So, the sum of the coefficients of each power of x must be zero. This implies that 2a2 + n ( n + 1)a0 = 0 for coefficients of x 0

(6.12)

6a3 + [ −2 + n( n + 1)]a1 = 0 for coefficients of x1

(6.13)

In general, when r = 2,3, ( r + 2)( r + 1)ar + 2 + [ − r ( r − 1) − 2r + n ( n + 1)]ar = 0

(6.14)

Simplifying the expression in square brackets, we obtain ( n − r ) ( n − r + 1) so that ar + 2 = −

Ch06.indd 5

( n − r )( n + r + 1) a r = 0,1,2,… ( r + 2)( r + 1) r

(6.15)

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6-6

Differential Equations

This is called a recurrence relation or recursion formula. It gives all the coefficients starting from a2 onwards in terms of a0 and a1 , which are considered as arbitrary constants. Thus we have n ( n + 1) a0 , 2! ( n − 1)( n + 2) a3 = − a1 , 3! ( n − 2)( n + 3) a4 = − a2 4⋅3 ( n − 2) n( n + 1)( n + 3) = a0 , 4! ( n − 3)( n + 4) a5 = − a3 5⋅ 4 ( n − 3)( n − 1)( n + 2)( n + 4) = a1 ,etc. 5! Substituting these coefficients in Eq. (6.9), we obtain a2 = −

y( x ) = a0 y1 ( x ) + a1 y2 ( x )

(6.16)

(6.17)

where y1 ( x ) = 1 − and

n( n + 1) 2 ( n − 2)n ( n + 1)( n + 3) 4 x + x − 2! 4!

( n − 1)( n + 2) 3 x 3! ( n − 3)( n − 1)( n + 2)( n + 4) 5 + x − 5!

(6.18)

y2 ( x ) = x −

(6.19)

The two series converge for x < 1, if they are non-terminating. Since y1 contains only the even powers of x and y2 contains only the odd powers of x, the ratio y1 y2 is not a constant so that y1 and y2 are linearly independent. Hence Eq. (6.17) is a general solution of Eq. (6.7) in the interval −1 < x < 1.

6.1.5

Legendre Polynomials Pn (x )

If the parameter n in Legendre’s Eq. (6.7) is a non-negative integer n, then the right-hand side of Eq. 6.15) is zero when r = n. This implies that an+ 2 = an+ 4 = = 0. Hence, if n is even, y1 ( x ) reduces to a polynomial of degree n, while y2 ( x ) remains an infinite series.

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Integration in Series

6-7

Similarly, if n is odd, y2 ( x ) reduces to a polynomial of degree n, while y1 ( x ) remains an infinite series. In either case, the terminating series solution (i.e., the polynomial solution) of the Legendre’s equation, multiplied by some constants are called Legendre polynomials or zonal harmonics of order n and are denoted by Pn ( x ). The series which is non-terminating is known as Legendre’s function of the second kind and is denoted by Qn ( x ). Thus, for a non-negative integer n the general solution of Legendre’s Eq. (6.7) is y( x ) = APn ( x ) + BQn ( x )

(6.20)

where Pn ( x ) is a polynomial of degree n and Qn ( x ) is a non-terminating series, which is unbounded at x = ±1. From Eq. (6.15), we have −( r + 2)( r + 1) (6.21) a for r ≤ ( n − 2) ( n − r )( n + r + 1) r + 2 Now, all the non-vanishing coefficients may be expressed in terms of the coefficient an of the highest power of x of the polynomial. The coefficient an which is still arbitrary may be chosen as an = 1 when n = 0 and ar =

an =

(2n)! 1 ⋅ 3 ⋅ 5 (2n − 1) = 2n ( n!)2 n! for n = 1, 2,…

(6.22)

For this choice of an , all these polynomials will have the value 1 when x = 1, i.e., Pn (1) = 1. We obtain from Eq. (6.21) and Eq. (6.22), n( n − 1) n( n − 1) (2n)! an = − ⋅ 2(2n − 1) 2(2n − 1) 2n ( n!)2 n( n − 1)2n(2n − 1)(2n − 2)! =− 2(2n − 1)2n n( n − 1)! n( n − 1)( n − 2)! (2n − 2)! =− n 2 ( n − 1)!( n − 2)!

an − 2 = −

i.e., an − 2

( n − 2)( n − 3) 4(2n − 3) (2n − 4)! = n and, in general, when ( n − 2r ) ≥ 0 2 2!( n − 2)!( n − 4)!

Similarly, an − 4 = − an − 2

an − 2 r = ( −1) r

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(2n − 2r )! 2 r !( n − r )!( n − 2r )! n

(6.23)

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6-8

Differential Equations

The resulting solution of Legendre’s differential equation Eq. (6.7) is called the Legendre polynomial of degree n, denoted by Pn ( x ) and is given by ( −1) r (2n − 2r )! x 2n−2r 2 r !( n − r )!( n − 2 r )! r=0 M

Pn ( x ) = ∑

n

(6.24)

where M = n 2 or ( n − 1) 2 according as n is even or odd so that M is an integer. Legendre’s equation Eq. (6.7) can also be solved as a series of descending powers of x, assuming a solution of Eq. (6.7) in the form ∞

y = ∑ ar x m − r r=0

Differentiating w.r.t. x twice, we obtain ∞

y ′ = ∑ ( m − r )ar x m − r −1 , r=0 ∞

y ′′ = ∑ ( m − r )( m − r − 1)ar x m − r − 2 r=0

Substituting these expressions in Eq. (6.7) ∞

∞

r=0 ∞

r=0

(1 − x 2 )∑ ar ( m − r )( m − r − 1) x m − r − 2 − 2 x ∑ ar ( m − r ) x m − r −1 + n( n + 1)

∑a x

∞

r=0

r

m−r

=0

⇒ ∑ ar [( m − r )( m − r − 1) x m − r − 2 + {n( n + 1) − ( m − r )( m − r − 1) r=0 ∞

−2( m − r )}x m − r ] = 0

⇒ ∑ ar [( m − r )( m − r − 1) x m − r − 2 + {n( n + 1) − ( m − r ) r=0 ∞

( m − r + 1)}x m − r ] = 0

⇒ ∑ ar [( m − r )( m − r − 1) x m − r − 2 + {n2 − ( m − r )2 + n r=0 ∞

−( m − r )}x m − r ] = 0

⇒ ∑ ar [( m − r )( m − r − 1) x m − r − 2 + ( n − m + r )( n + m − r + 1) x m − r ] = 0 r=0

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Integration in Series

6-9

Equating to zero the coefficient of the highest power of x, i.e., x m , we get a0 ( n − m)( n + m + 1) = 0 which is obtained by putting r = 0 in the coefficient of x m − r in (6.9). This implies that m = n or m = − n − 1 (∵ a0 ≠ 0) Equating now to zero the coefficient of the next power of x, i.e., x m −1 , we get a1 ( n − m + 1)( n + m) = 0 which is obtained by putting r = 1 in the coefficient of x m − r −1 in Eq. (6.25). This implies that a1 = 0 ∵ m ≠ − n and m ≠ n ÷ 1 To obtain the recurrence relation, we equate the coefficient of x m − r to zero: ( m − r + 2)( m − r + 1)ar − 2 − ( m − r − n)( m − r + n + 1)ar = 0 ( m − r + 2)( m − r + 1) ⇒ ar = − a r = 2,3, 4,… ( m − r + n)( m − r + n + 1) r − 2 From this relation, we see that a1 = a3 = a5 =

= a2 r −1 = 0.

Case (i) m = n, We have ar =

( n − r + 2)( n − r + 1) ar − 2 r (2n − r + 1)

Putting r = 2, 4,6,… , we have n( n − 1) a 2(2n − 1) 0 ( n − 2)( n − 3) a4 = − a 4(2n − 3) 2 n( n − 1)( n − 2)( n − 3) a = 2 ⋅ 4 ⋅ (2n − 1)(2n − 3) 0 ( n − 4)( n − 5) a6 = − a4 6(2n − 5) n( n − 1)( n − 2)( n − 3)( n − 4)( n − 5) a ,etc. =− 2 ⋅ 4 ⋅ 6 ⋅ (2n − 1)(2n − 3)(2n − 5) 0 a2 = −

We obtain a solution of Eq. (6.7) as ⎡ n( n − 1) n − 2 n( n − 1)( n − 2)( n − 3) n − 4 y1 = a0 ⎢ x n − x + x − 2(2n − 1) 2 ⋅ 4 ⋅ (2n − 1)(2n − 3) ⎣

Ch06.indd 9

⎤ ⎥ ⎦

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6-10

Differential Equations

(2n − 1) the first solution y1 is n! obtained and is called Legendre’s polynomial and is denoted by Pn ( x ). This is called the Legendre’s function of the first kind of degree n. If we choose a0 as a0 =

1⋅ 3 ⋅ 5

Case (ii) m = − n − 1, In this case, ar = Putting

( n + r − 1)( n + r ) a r (2n + r + 1) r − 2

r = 2, 4, 6 ( n + 1)( n + 2) a 2(2n + 3) 0 ( n + 3)( n + 4) ( n + 1)( n + 2)( n + 3)( n + 4) a4 = a2 = a0 , 4(2n + 5) 2 ⋅ 4 ⋅ (2n + 3)(2n + 5) a2 =

∴

∞ ⎡ ( n + 1)( n + 2) − n −3 y2 = ∑ ar x − n −1− r = a0 ⎢ x − n −1 + x 2(2n + 3) r=0 ⎣ ⎤ ( n + 1)( n + 2)( n + 3)( n + 4) − n −5 x + + ⎥ 2 ⋅ 4 ⋅ (2n + 3)(2n + 5) ⎦

Choosing a0 = 1⋅3⋅5 n(2! n+1) the second solution y2 of Legendre’s Eq. (6.7) is obtained and it is called Legendre’s function of the second kind of degree n and is denoted by Qn ( x ). The general solution of Legendre’s Eq. (6.7) is y = APn ( x ) + BQn ( x ), where A and B are arbitrary constants. The first few Legendre polynomials are shown in Fig. 6.1. ( n = 0, M = 0) P0 ( x ) = 1 ( n = 1, M = 0) P1 ( x ) = x 1 P2 ( x ) = (3 x 2 − 1) ( n = 2, M = 1, r = 0 to 1) 2 1 P3 ( x ) = (5 x 3 − 3 x ) ( n = 3, M = 1, r = 0 to 1) 2 1 P4 ( x ) = (35 x 4 − 30 x 2 + 3)( n = 4, M = 2, r = 0 to 2) 8 1 P5 ( x ) = (63 x 5 − 70 x 3 + 15 x )( n = 5, M = 2, r = 0 to 2) 8 1 P6 ( x ) = [231x 6 − 315 x 4 + 105 x 2 − 5]( n = 6, M = 3, r = 0 to 3) (6.25) 16

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Integration in Series

6-11

Pn (x ) P0

1

P1 P4 P3

–1

1 x

P2 –1

Figure 6.1 Legendre Polynomials Rodrigues’3 Formula for Legendre Polynomials Pn ( x ) Example 6.1.1 Show that Pn ( x ) =

1 dn [( x 2 − 1) n ] ( n ∈ ) (6.26). n!2n dx n

Solution Let y = ( x 2 − 1) n . Then dy = n( x 2 − 1) n −1 2 x dx dy ⇒ − (1 − x 2 ) = 2nx( x 2 − 1) n = 2nxy dx dy ⇒ (1 − x 2 ) + 2nxy = 0 (6.27) dx Differentiating Eq. (6.27) ( n + 1) times using Leibnitz’s theorem, we have y1 =

(1 − x 2 ) yn+ 2 + ( n + 1)( −2 x ) yn+1 − ( n + 1)nyn + 2n [ xyn+1 + ( n + 1) ⋅ 1 ⋅ yn ] = 0 d 2v dv − 2 x + n( n + 1)v = 0 (where v = yn ) 2 dx dx which is Legendre’s differential equation having Pn ( x ) and Qn ( x ) as solutions. or

(1 − x 2 )

3

Olinde Rodrigues (1794–1851) was a French mathematician and economist.

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6-12

Differential Equations

dn [( x 2 − 1) n ] contains positive powers of x only, n dx v must be a constant multiple of Pn ( x ). dn Let Pn ( x ) = Cv = Cyn = C n ( x 2 − 1) n = CD n [( x − 1) n ( x + 1) n ] dx = C [ D n ( x − 1) n ⋅ ( x + 1) n + n C1 ⋅ D n −1 ( x − 1) n ⋅ ( x + 1) n −1 + + ( x − 1) n ⋅ D n ( x + 1) n ] = C [n!( x + 1) n + terms containing ( x − 1) as a factor] Since v = yn =

Putting x = 1, we get 1 = Pn (1) = C ( n!2n ) ⇒ C = ∴ Pn ( x ) =

1 n!2n

1 dn [( x 2 − 1) n ] n!2n dx n

Note 1. The first few Legendre polynomials obtained earlier can also be obtained using Rodrigues’ formula. 2. Any polynomial f ( x ) of degree n can be expressed as a linear combination of Legendre polynomials Pn ( x ) as n

f ( x ) = ∑ Cm Pm ( x ) m= 0

6.1.6 Generating Function for Legendre Polynomials Pn ( x ) Generating function: Let < f n ( x ) > be a sequence of functions. A function w (t , x ) is called a generating function for the functions f n ( x ) if w (t , x ) =

∞

∑

n = −∞

f n ( x )t n .

Example 6.1.2 Show that (1 − 2 xt + t 2 )

−

1 2

∞

= ∑ n= 0 t n Pn ( x ).

Solution We have, by binomial theorem, 1 1⋅ 3 2 1⋅ 3 ⋅ 5 3 1 1 u + u +… (1 − 2 xt + t 2 ) − 2 = (1 − u ) − 2 = 1 + u + 2 2⋅4 2⋅4⋅6 1 ⋅ 3 ⋅ 5 (2m − 1) m u + + 2 ⋅ 4 ⋅ 6 (2m)

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Integration in Series

6-13

where u = 2 xt − t 2 . ∞

1 ⋅ 3 ⋅ 5 (2m − 1) (2 xt − t 2 ) m (2m) m =1 2 ⋅ 4 ⋅ 6

= 1+ ∑

u m = t m (2 x − t ) m = t m [(2 x ) m − m(2 x ) m −1 m( m − 1) + ⋅ (2 x ) m − 2 ⋅ t 2 + 2! Substituting in Eq. (6.28), we get

(6.28)

Now,

(1 − 2 xt + t 2 )

−

1 2

+ ( −1) m t m ]

∞

1 ⋅ 3 ⋅ 5 (2m − 1) [(2 x ) − m(2 x ) m −1 ⋅ t m+1 2 ⋅ 4 ⋅ 6 (2 m ) m =1 m( m − 1) + ⋅ (2 x ) m − 2 t m+ 2 + + ( −1) m t 2 m (6.29) 2!

= 1+ ∑

Thus, the coefficient of t n in the expansion (1 − 2 xt + t 2 ) − 2 is 1

1 ⋅ 3 ⋅ (2n − 1) 1 ⋅ 3 (2n − 3) (2 x ) n − ( n − 1)(2 x ) n 2 ⋅ 4 (2n) 2 ⋅ 4 (2n − 2) 1 ⋅ 3 (2n − 5) ( n − 2)( n − 3) + (2 x ) n − 4 + 2 ⋅ 4 (2n − 4) 2! 1 ⋅ 3 ⋅ 5 (2n − 1) ⎡ n n( n − 1) n − 2 = x − x ⎢ n! 2 ⋅ (2n − 1) ⎣ ⎤ n( n − 1)( n − 2)( n − 3) n − 4 x − ⎥ = Pn ( x ) + 2 ⋅ 4 ⋅ (2n − 1)(2n − 3) ⎦ ∴

∞

(1 − 2 xt + t 2 ) −1 2 = ∑ t n Pn ( x )

(6.30)

n= 0

6.1.7

Recurrence Relations of Legendre Functions

Example 6.1.3 Show that the following recurrence relations a satisfied by the Legendre polynomials Pn ( x ) : RR1: (2n + 1) xPn ( x ) = ( n + 1) Pn+1 ( x ) + nPn −1 ( x ) RR2: nPn ( x ) = xPn′( x ) − Pn′−1 ( x ) RR3: (2n + 1) Pn ( x ) = Pn′+1 ( x ) − Pn′−1 ( x ) RR4: Pn′( x ) = xPn′−1 ( x ) + nPn −1 ( x )

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6-14

Differential Equations

RR5: (1 − x 2 ) Pn′( x ) = n[ Pn −1 ( x ) − xPn ( x )] RR6: (1 − x 2 ) Pn′( x ) = ( n + 1)[ xPn ( x ) − Pn+1 ( x )] Solution To prove RR1 ∞

We know that (1 − 2 xt + t 2 ) −1 2 = ∑ t n Pn ( x ).

(6.31)

n= 0

Differentiating Eq. (6.31) partially w.r.t. t, we get ∞ 1 − (1 − 2 xt + t 2 ) −3 2 ( −2 x + 2t ) = ∑ nt n −1 Pn ( x ). 2 n =1

Multiplying this by (1 − 2 xt + t 2 ) and using Eq. (6.31), we get ∞

∞

( x − t )∑ t n Pn ( x ) = (1 − 2 xt + t 2 )∑ nt n −1 Pn ( x ) n= 0

n =1

n

Equating the coefficients of t xPn ( x ) − Pn −1 ( x ) = ( n + 1) Pn+1 ( x ) − 2nxPn ( x ) + ( n − 1) Pn −1 ( x ) ⇒ (2n + 1) xPn ( x ) = ( n + 1) Pn +1 ( x ) + nPn −1 ( x ) To prove RR2 nPn ( x ) = xPn′( x ) − Pn′−1 ( x ) Differentiating Eq. (6.31) partially w.r.t. t and x, we get respectively ( x − t )(1 − 2 xt + t 2 ) −3 2

∞

= ∑ nt n −1 Pn ( x ) n =1

and

t (1 − 2 xt + t 2 ) −3 2

∞

= ∑ t n Pn′( x ). n= 0

From these we have, on multiplying the first by t and the second by ( x − t ), ∞

∞

n =1

n= 0

t ( x − t )(1 − 2 xt + t 2 ) −3 2 = ∑ nt n Pn ( x ) = ( x − t )∑ t n Pn′( x ) n

Equating the coefficients of t nPn ( x ) = xPn′( x ) − Pn′−1 ( x )

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Integration in Series

6-15

To prove RR3 (2n + 1) Pn ( x ) = Pn′+1 ( x ) − Pn′−1 ( x ) Differentiating w.r.t. x we have

RR1:

(2n + 1) xPn ( x ) = ( n + 1) Pn+1 ( x ) + nPn −1 ( x )

(2n + 1) xPn′( n) + (2n + 1) Pn ( x ) = ( n + 1) Pn′+1 ( x ) + nPn′−1 ( x ) RR2 is xPn′( x ) = nPn ( x ) + Pn′−1 ( x ) ∴ (2n + 1)[nPn ( x ) + Pn′−1 ( x )] + (2n + 1) Pn ( x ) = ( n + 1) Pn′+1 ( x ) + nPn′−1 ( x ). ⇒ (2n + 1)( n + 1) Pn ( x ) = ( n + 1) Pn′+1 ( x ) − ( n + 1) Pn′−1 ( x ). ⇒ (2n + 1) Pn ( x ) = Pn′+1 − Pn′−1 ( x ), cancelling ( n + 1) ≠ 0.

To prove RR4 Pn′( x ) = xPn′−1 ( x ) + nPn −1 ( x ) RR3: (2n + 1) Pn ( x ) = Pn′+1 ( x ) − Pn′−1 ( x ) RR2: nPn ( x ) = xPn′( x ) − Pn′−1 ( x ) Subtracting we get ( n + 1) Pn ( x ) = Pn′+1 ( x ) − xPn′( x ) Replace ( n + 1) by n nPn −1 ( x ) = Pn′( x ) − xPn′−1 ( x ) or

Pn′( x ) = nPn −1 ( x ) + xPn′−1 ( x )

To prove RR5 (1 − x 2 ) Pn′( x ) = n[ Pn −1 ( x ) − xPn ( x )] RR4: Pn′( x ) − xPn′−1 ( x ) = nPn −1 ( x ) RR2: xPn′( x ) − Pn′−1 ( x ) = nPn ( x ) Multiplying the last equation by x and subtracting it from the previous one (1 − x 2 ) Pn′( x ) = n[ Pn −1 ( x ) − xPn ( x )]

Ch06.indd 15

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6-16

Differential Equations

To prove RR6 (1 − x 2 ) Pn′( x ) = ( n + 1)[ xPn ( x ) − Pn+1 ( x )] Rewriting RR1 as ( n + 1) xPn ( x ) + nxPn ( x ) = ( n + 1) Pn+1 ( x ) + nPn −1 ( x ) ⇒ ( n + 1)[ xPn ( x ) − Pn +1 ( x )] = n[ Pn −1 ( x ) − xPn ( x )] = (1 − x 2 ) Pn′( x ) by RR5 2 ⇒ (1 − x ) Pn′( x ) = ( n + 1)[ xPn ( x ) − Pn+1 ( x )] Note The above recurrence relations can also be proved by using Rodrigues’ formula. Example 6.1.4 (i) Pn (1) = 1;

(iii) Pn ( −1) n .

(ii) Pn ( − x ) = ( −1) n Pn ( x );

Solution We know that ∞

∑t

n

n= 0

Pn ( x ) = (1 − 2 xt + t 2 ) − 2 1

(6.32)

(i) Put x = 1 in Eq. (6.32) ∞

∑t n= 0

n

Pn (1) = (1 − 2t + t 2 ) − 2 = [(1 − t )2 ]− 2 = (1 − t ) −1 1

= 1+ t + t2 + t3 +

1

∞

= ∑ t n ⋅1 n= 0

Equating the coefficients of t on both sides Pn(1) = 1. (ii) Replacing t by –t and x by –x in Eq. (6.32), we get n

∞

∑ ( −1) t

n n

n= 0

Pn ( − x ) = [1 − 2( − x )( −t ) + ( −t )2 ]− 2 1

∞

= (1 − 2 xt + t 2 ) − 2 = ∑ t n Pn ( x ) 1

n= 0

n

Equating the coefficients of t on both sides ( −1) n Pn ( − x ) = Pn ( x )

or

Pn ( − x ) = ( −1) n Pn ( x ).

(iii) Put x = 1 in the above result. Then, Pn ( −1) = ( −1) n . ∵ Pn (1) = 1

Ch06.indd 16

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Integration in Series

6.1.8

6-17

Orthogonality of Functions

A set of functions f1 , f 2 , f 3 ,… defined on some interval I = {x ∈ R | a ≤ x ≤ b} is said to be orthogonal on I with respect to a weight function p( x ) > 0 if

∫

b

a

p( x ) f m ( x ) f n ( x )dx = 0

for m ≠ n

(6.33)

The norm f n of f n is defined by fn =

∫

b

a

p( x ) f n2 ( x )dx

(6.34)

The functions are called orthonormal on I if they are orthogonal on I and all have a norm equal to 1. In respect of functions with p( x ) = 1, we simply say ‘orthogonal’. Thus, functions f1 , f 2 , f 3 ,… are orthogonal on some interval I if

∫

b

a

f m ( x ) f n ( x )dx = 0

m≠n

for

(6.35)

The norm f n of f n is then simply given by fn =

∫

b

a

f n2 ( x )dx

(6.36)

The functions are called orthogonal on I if they are orthogonal on I and all of them are of norm equal to 1.

6.1.9

Orthogonality of Legendre Polynomials Pn ( x )

Example 6.1.5 Show that the Legendre Polynomials P0 ( x ), P1 ( x ), P2 ( x ),… are orthogonal on I = [ −1,1]. That is, for m ≠ n ⎧0 ⎪ ∫−1 Pm ( x) Pn ( x)dx = ⎨ 2 for m = n ⎪⎩ 2n + 1 1

Solution Case (i) m ≠ n

(6.37)

Let u = Pm ( x ) and v = Pn ( x ) be the solutions of the Legendre’s equation (1 − x 2 ) y ′′ − 2 xy ′ + p( p + 1) y = 0

Ch06.indd 17

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6-18

Differential Equations

so that we have (1 − x 2 )u ′′ − 2 xu ′ + m( m + 1)u = 0

(6.38)

(1 − x 2 )v ′′ − 2 xv ′ + n( n + 1)v = 0

(6.39)

Multiplying Eq. (6.38) by v and Eq. (6.39) by u and subtracting, we have (1 − x 2 )(u ′′v − uv ′′ ) − 2 x(u ′v − uv ′ ) + [m( m + 1) − n( n + 1)]uv = 0 (1 − x 2 )(u ′v − uv ′ ) ′ + (1 − x 2 )′ (u ′v − uv ′ ) + ( m − n)( m + n + 1)uv = 0 [(1 − x 2 )(u ′v − uv ′ ) ′ + ( m − n)( m + n + 1)uv = 0 On transposing, we have − ( m − n)( m + n + 1)uv = [(1 − x 2 )(u ′v − uv ′ )]′ Integrating both sides w.r.t. x between –1 and 1, we get 1

− ( m − n)( m + n + 1) ∫ uvdx = (1 − x 2 )(u ′v − uv ′ ) −1 = 0 m ≠ n 1

−1 1

⇒ ∫ uvdx = 0 or −1

∫

1

−1

Pm ( x ) Pn ( x )dx = 0( m ≠ n).

(6.40)

Case (ii) m = n. We know that

∑t

n

Pn ( x ) = P0 ( x ) + tP1 ( x ) + t 2 P2 ( x ) +

= (1 − 2 xt + t 2 ) − 2 1

(6.41)

Squaring both sides we have P02 ( x ) + t 2 P12 ( x ) + t 4 P22 ( x ) + +2[tP0 ( x ) P1 ( x ) + t 2 ( P0 ( x ) P2 ( x )) + t 3 ( P0 ( x ) P2 ( x ) + P1 ( x ) P2 ( x )) + ] = (1 − 2 xt + t 2 ) −1 ∞

∑t n= 0

2n

Pn2 ( x ) + 2

∞

∑

m =1( m ≠ n )

(6.42)

t m+ n Pm ( x ) Pn ( x ) = (1 − 2 xt + t 2 ) −1 (6.43)

Integrating both sides w.r.t. x between –1 and 1, we get ∞

∑t ∫ 2n

n= 0

Ch06.indd 18

1

−1

Pn2 ( x )dx + 0 = ∫

1

−1

dx 1 − 2 xt + t 2

(6.44)

12/13/2011 11:11:53 AM

Integration in Series

6-19

by Eq. (6.40). 1

⎡ 1 ⎤ = ⎢ − log(1 − 2 xt + t 2 ) ⎥ t 2 ⎣ ⎦ −1 1 = − [log(1 − t )2 − log(1 + t )2 ] 2t 1 = − [log(1 − t ) − log(1 + t )] t 1 ⎡ t2 t3 ⎛ t2 t3 ⎞⎤ = ⎢t + + + + ⎜ t − + − ⎟ ⎥ ⎝ ⎠⎦ t⎣ 2 3 2 3 ∞ 3 4 2n t t t 2n ⎡ t ⎤ = 2 ⎢1 + + + + + ⎥ = 2∑ 2n + 1 ⎣ 3 5 ⎦ n = 0 2n + 1 2n Equating the coefficients of t on both sides, we get

∫

1

−1

Pn2 ( x )dx =

2 . 2n + 1

(6.45)

(6.46)

Example 6.1.6 Prove that Pn′+1 ( x ) + Pn′( x ) = P0 ( x ) + 3P1 ( x ) + 5 P2 ( x ) + + (2n + 1) Pn ( x ) n

= ∑ (2m + 1) Pm ( x ).

(6.47)

m= 0

Solution RR3 ⇒ (2n + 1) Pn = Pn′+1 − Pn′−1 Replacing n by 1,2,… ( n − 1), n 3P1 = P2′ − P0′; 5 P2 = P3′ − P1′; 7 P3 = P4′ − P2′(2n − 1) Pn −1 = Pn′ − Pn′− 2 (2n + 1) Pn = Pn′+1 − Pn′−1 Adding we get 3P1 + 5 P2 + 7 P3 + + (2n − 1) Pn −1 + (2n + 1) Pn = − P0′ − P1′+ Pn′ + Pn′+1 n

⇒ Pn′+1 ( x ) + Pn′( x ) = ∑ (2m + 1) Pm ( x ) m= 0

(∵ − P1′ = − P0 and P0′ = 0)

Ch06.indd 19

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6-20

Differential Equations

6.1.10 Betrami’s Result Example 6.1.7 Prove that (2n + 1)( x 2 − 1) Pn′( x ) = n( n + 1)[ Pn+1 ( x ) − Pn −1 ( x )]. Solution RR5: (1 − x 2 ) Pn′( x ) = n[ Pn −1 ( x ) − xPn ( x )]

(6.48)

RR6: (1 − x 2 ) Pn′( x ) = ( n + 1)[ xPn ( x ) − Pn+1 ( x )]

(6.49)

Multiplying Eq. (6.48) by ( n + 1), Eq. (6.49) by n and adding [( n + 1) + n](1 − x 2 ) Pn′( x ) = n( n + 1)[ Pn −1 ( x ) − xPn ( x ) + xPn ( x ) − Pn+1 ( x )] ⇒ (2n + 1)( x 2 − 1) Pn′( x ) = n( n + 1)[ Pn+1 ( x ) − Pn −1 ( x )]

6.1.11 Christoffel’s Expansion Example 6.1.8 Prove that Pn′ = (2n − 1) Pn −1 + (2n − 5) Pn −3 + (2n − 9) Pn −5 + + 3P1 or P0 according as n is even or odd. Solution RR3: (2n + 1) Pn = Pn′+1 − Pn′−1 ⇒ Pn′+1 = (2n + 1) Pn + Pn′−1 Replacing n by (n – 1), we have Pn′ = (2n − 1) Pn −1 + Pn′− 2

(6.50)

Replacing n by (n – 2), (n – 4), (n – 6),… and finally by 2 (if n is even) and 3 (if n is odd).

Ch06.indd 20

Pn′− 2 = (2n − 5) Pn −3 + Pn′− 4

(6.51)

Pn′− 4 = (2n − 9) Pn −5 + Pn′−6

(6.52)

P2′ = 3P1 + P0′

( n even)

(6.53)

P3′ = 5 P2 + P ′

( n odd)

(6.54)

12/13/2011 11:11:54 AM

Integration in Series

6-21

Adding these Pn′ = (2n − 1) Pn −1 + (2n − 5) Pn −3 + (2n − 9) Pn −5 + + 3P1 ( n even) ∵ P0′ = 0 = (2n − 1) Pn −1 + (2n − 5) Pn −3 + (2n − 9) Pn −5 + + P1′ ( n odd)

6.1.12 Christoffel’s Summation Formula n

Example 6.1.9 Prove that

∑ (2m + 1) P ( x) P ( y) m

m= 0

= ( n + 1) ×

m

[ Pn+1 ( x ) Pn ( y ) − Pn +1 ( y ) Pn ( x )] ( x − y)

( x ≠ y ).

Solution RR1 ⇒ (2m + 1) xPm ( x ) = ( m + 1) Pm+1 ( x ) + mPm −1 ( x )

(6.55)

(2m + 1) yPm ( y ) = ( m + 1) Pm+1 ( y ) + mPm −1 ( y )

(6.56)

(1) ⋅ Pm ( y ) − (2) ⋅ Pm ( x ) yields (2m + 1)( x − y ) Pm ( x ) Pm ( y ) = ( m + 1) × [ Pm+1 ( x ) Pm ( y ) − Pm+1 ( y ) Pm ( x )] − m[ Pm −1 ( y ) Pm ( x ) − Pm −1 ( x ) Pm ( y )] (6.57) Substituting m = 0,1, 2,… n in Eq. (6.57) 1 ⋅ ( x − y ) P0 ( x ) P0 ( y ) = 1[ P1 ( x ) P0 ( y ) − P1 ( y ) P0 ( x )]

(6.58)

3 ⋅ ( x − y ) P1 ( x ) P1 ( y ) = 2 ⋅ [ P2 ( x ) P1 ( y ) − P2 ( y ) P1 ( x )] −1 ⋅ [ P1 ( x ) P0 ( y ) − P1 ( y ) P0 ( x )]

(6.59)

5 ⋅ ( x − y ) P2 ( x ) P2 ( y ) = 3 ⋅ [ P3 ( x ) P2 ( y ) − P3 ( y ) P2 ( x )] −2 ⋅ [ P2 ( x ) P1 ( y ) − P2 ( y ) P1 ( x )]

(6.60)

(2n − 1)( x − y ) Pn −1 ( x ) Pn −1 ( y ) = n[ Pn ( x ) Pn −1 ( y ) − Pn ( y ) Pn −1 ( x )] −( n − 1)[ Pn −1 ( x ) Pn − 2 ( y ) − Pn −1 ( y ) Pn − 2 ( x )] (6.61) (2n + 1)( x − y ) Pn ( x ) Pn ( y ) = ( n + 1) × [ Pn+1 ( x ) Pn ( y ) − Pn+1 ( y ) Pn ( x )] − n[ Pn ( x ) Pn −1 ( y ) − Pn ( y ) Pn −1 ( x )] (6.62)

Ch06.indd 21

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6-22

Differential Equations

Adding Eqs. (6.58) – (6.62) n

( x − y ) ∑ (2m + 1) Pm ( x ) Pm ( y ) m= 0

= ( n + 1) × [ Pn +1 ( x ) Pn ( y ) − Pn +1 ( y ) Pn ( x )]

n

⇒ ∑ (2m + 1) Pm ( x ) Pm ( y ) = ( n + 1) m= 0

×

6.1.13

[ Pn +1 ( x ) Pn ( y ) − Pn +1 ( y ) Pn ( x )] ( x − y)

( x ≠ y)

Laplace’s First Integral for Pn ( x )

Example 6.1.10 Show that when n ∈ , Pn ( x ) =

π

1

π∫

0

( x ± x 2 − 1cos θ ) n dθ .

Solution We know that π dθ ∫0 a ± b cos θ =

π a −b 2

( a2 > b2 )

2

(6.63)

Let a = 1 − tx and b = t x 2 − 1. We have a 2 − b 2 = 1 − 2tx + t 2 so that Eq. (6.63) becomes

∫

π

0

∞

(1 − tx ± t x 2 − 1cos θ ) −1 dθ = π (1 − 2tx + t 2 ) −1 2 = π ∑ t n Pn ( x ) n= 0

[∵ (1 − 2 xt + t 2 )

− 12

is the generating function for Pn ( x )].

∞

π

n= 0

0

⇒ π ∑ t n Pn ( x ) = ∫ (1 − tu ) −1 dθ [where u = x ± x 2 − 1cos θ ] =∫

π

π ∞

0

π

∑ (tu)n dθ = ∑ t n ∫ ( x ± x 2 − 1cosθ )n dθ n= 0

n= 0

0

Equating the coefficients of t n,

( 1 P ( x) = ∫ ( x ± π π

) − 1cos θ ) dθ n

π Pn ( x ) = ∫ x ± x 2 − 1cos θ dθ 0

or

Ch06.indd 22

π

n

0

x2

n

12/13/2011 11:11:55 AM

Integration in Series

6-23

Example 6.1.11 Show that 1

π∫

Pn (cos φ ) =

π

0

(cos φ + i sin φ cos θ ) n dθ .

Solution Laplace’s first integral for Pn ( x ) is Pn ( x ) = Put x = cos φ then

1

π∫

6.1.14

0

(x ±

)

n

x 2 − 1cos θ dθ

x 2 − 1 = cos 2 φ − 1 = −1 1 − cos 2 φ = i sin φ

Pn (cos φ ) =

∴

π

1

π∫

π

0

(cos φ + i sin φ cos φ )dθ

Laplace’s Second Integral for Pn ( x )

Example 6.1.12 Prove that 1 π Pn ( x ) = ∫ x ± x 2 − 1cos θ

π

0

(

)

− n −1

dθ ( n ∈ ).

Solution We know that

∫

π

0

dθ = a ± b cos θ

π a − b2 2

( a2 > b2 )

Let a = tx − 1 and b = t x 2 − 1. Then 1 ⎞ ⎛1 a 2 − b 2 = 1 − 2 xt + t 2 = t 2 ⎜ 2 − 2 x + 1⎟ ⎝t t ⎠ ∴

−1 2

π⎛

π 1 1⎞ = ∫ ( −1 + tx ± t x 2 − 1cos θ ) −1 dθ ⎜⎝1 − 2 x + 2 ⎟⎠ 0 t t t n ∞ π π ⎛ 1⎞ ⇒ ∑ ⎜ ⎟ Pn ( x ) = ∫ (u − 1) −1 dθ 0 t n= 0 ⎝ t ⎠

where u = t [ x ± x 2 − 1cos θ ] −1

π 1⎛ 1 ⎛ 1⎞ 1 1 ⎞ = ∫ ⎜1 − ⎟ dθ = ∫ ⎜1 + + 2 + ⎟ dθ 0 u⎝ 0 ⎠ u⎠ u⎝ u u ∞ π⎛ ∞ π dθ 1 ⎞ = ∫ ⎜ ∑ n +1 ⎟ dθ = ∑ ∫ 0 ⎝ 0 n +1 ⎠ t ( x ± x 2 − 1cos θ ) n +1 n= 0 u n= 0 π

Ch06.indd 23

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6-24

Differential Equations

Equating the coefficients of

π Pn ( x ) = ∫

π

0

or

6.1.15

Pn ( x ) =

1

π∫

1 t

n +1

on both sides

dθ ( x ± x − 1cos θ ) n +1 2

dθ ∵ P− ( n+1) = Pn ( x ) ( x ± x − 1cos θ ) n+1

π

2

0

Expansion of f(x) in a Series of Legendre Polynomials

Example 6.1.13 Let f (x) be expressible in a series of Legendre polynomials. Thus ∞

f ( x ) = ∑ Cn Pn ( x ) = C0 P0 ( x ) + C1 P1 ( x ) + n= 0

+ Cn Pn ( x ) +

(6.64)

where Cn are constants to be determined. Then multiplying both sides by Pn ( x ) and integrating w.r.t. x from –1 to 1, we get

∫

1

1

f ( x ) Pn ( x )dx = ∫ [C0 P0 ( x ) + C1 P1 ( x ) + + Cn Pn ( x ) + −1 −1 1 2 = Cn ∫ Pn2 ( x )dx = Cn ⋅ −1 2n + 1 2n + 1 1 ⇒ Cn = f ( x ) Pn ( x )dx 2 ∫−1 1

∵ ∫ Pm ( x ) Pn ( x )dx = 0 −1 1 2 ∫−1 Pn2 ( x)dx = 2n + 1

]Pn ( x )dx

if m ≠ n if m = n

Example 6.1.14 Show that P0 ( x ) = 1, P1 ( x ) = x, P2 ( x ) = 12 (3 x 2 − 1) and hence express 2 x 2 − 4 x + 2 in terms of Legendre polynomials. JNTU 2003S] Solution From Rodrigues’ formula Pn ( x ) =

Ch06.indd 24

1 dn 2 ( x − 1) n 2n n! dx n

12/13/2011 11:11:56 AM

Integration in Series

6-25

Taking n = 0, 1, 2 we have d0 2 ( x − 1)0 = 1 dx 0 1 d 2 ( x − 1) = x P1 ( x ) = 1 2 1! dx 1 d2 2 ( x − 1)2 P2 ( x ) = 2 2 2! dx 2 1 d 1 d 3 1 [2( x 2 − 1) ⋅ 2 x ] = ( x − x ) = (3x 2 − 1) ⇒ P2 ( x ) = 8 dx 2 dx 2 1 1 ⇒ x 2 = [2 P2 ( x ) + 1] = [2 P2 ( x ) + P0 ( x )]∵ P0 ( x ) = 1 3 3 2 4 2 ∴ 2 x − 4 x + 2 = [2 P2 ( x ) + P0 ( x )] − 4 P1 ( x ) + 2 P0 ( x ) = [ P2 ( x ) 3 3 −3P1 ( x ) + 2 P0 ( x )] P0 ( x ) =

Example 6.1.15 Express x 3 + 2 x 2 − x − 3 in terms if Legendre polynomials. Solution We have P0 ( x ) = 1, P1 ( x ) = x, 1 1 P2 ( x ) = (3 x 2 − 1), P3 ( x ) = (5 x 3 − 3 x ) 2 2 2 3 2 3 3 ⇒ x = P3 ( x ) + x = P3 ( x ) + P1 ( x ) 5 5 5 5 1 x 2 = [2 P2 ( x ) + P0 ( x )] x = P1 ( x ) 3 2 3 4 3 ∴ x + 2 x 2 − x − 3 = P3 ( x ) + P1 ( x ) + P2 ( x ) 5 5 3 2 + P0 ( x ) − P1 ( x ) − 3P0 ( x ) 3 2 4 2 7 = P3 ( x ) + P2 ( x ) − P1 ( x ) − P0 ( x ) 5 3 5 3 Example 6.1.16 Using Rodrigues’ formula prove that

∫

1

−1

x m Pn ( x )dx = 0.

if m < n

Solution Rodrigues’ formula for Pn ( x ) is Pn ( x ) =

Ch06.indd 25

1 dn 2 ( x − 1) n 2n n! dx n

12/13/2011 11:11:56 AM

6-26

Differential Equations

1 1 m dn 2 x ( x − 1) n dx 2 n!∫−1 dx n 1 1 = n ∫ x m D n ( x 2 − 1) n dx 2 n! −1 dn where D n = n dx 1 1 m = n ∫ x D[ D n −1 ( x 2 − 1) n ]dx 2 n! −1 1 = n [ x m D n −1 ( x 2 − 1) n ]1−1 2 n! 1 − ∫ mx m −1 D n −1 ( x 2 − 1) n dx −1 (by integration by parts) m 1 = 0 − n ∫ x m −1 D n −1 ( x 2 − 1) n dx 2 n! −1

1

⇒ ∫ x m Pn ( x )dx = −1

n

(∵ D n −1 ( x 2 − 1) n contains ( x 2 − 1) as a factor) Proceeding similarly, integrating by parts ( n − 1) times, we get 1 ( −1) n m 1 2 m x P ( x ) dx = ( x − 1) n D n −1 ( x m −1 )dx = 0 ∫−1 n 2n n! ∫−1 [∵ D m −1 ( x m −1 ) = 0 if m < n] 1

⇒ ∫ x m Pn ( x )dx = 0

if m < n.

−1

Example 6.1.17 Prove that (1 − 2 xt + t 2 ) −1 2 is a solution of the equation t

∂ 2 (tu ) ∂ ⎡ ∂u ⎤ + ⎢(1 − x 2 ) ⎥ = 0. 2 ∂t ∂x ⎣ ∂x ⎦

Solution We know that ∞

(1 − 2 xt + t 2 ) −1 2 = ∑ t n Pn ( x ) n= 0

Denoting each side by u, we have ∞

tu = ∑ t n+1 Pn . n= 0

⇒t

Ch06.indd 26

∂ (tu ) ∂u ∞ n n = n ( n + 1) t P and = ∑t P′ ∑ n ∂t 2 ∂x n = 0 n n= 0 2

∞

(6.65)

12/13/2011 11:11:56 AM

Integration in Series

∴

∞ ⎤ ∂ ⎡ ∂u ⎤ ∂ ⎡ 2 2 (1 ) (1 ) x x t n Pn′⎥ − = − ∑ ⎢ ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂x ⎣ n= 0 ⎦ ∞

∞

n= 0

n= 0

= (1 − x 2 )∑ t n Pn′′ − 2 x ∑ t n Pn′

6-27

(6.66)

From Eqs. (6.65) and (6.66) we have t

∂ 2 (tu ) ∂ ⎡ ∂u ⎤ ∞ 2 + (1 − x ) = ∑ n( n + 1)t n Pn ∂t 2 ∂x ⎢⎣ ∂x ⎥⎦ n= 0 ∞

∞

n= 0

n= 0

+(1 − x 2 ) ⋅ ∑ t n Pn′′ − 2 x ∑ t n Pn′ ∞

= ∑ t n [n( n + 1) Pn − 2 xPn′ n= 0

+(1 − x 2 ) Pn′′ ] = 0, since Pn satisfies Legendre’s equation: (1 − x 2 ) y ′′ − 2 xy ′ + n( n + 1) y = 0. Example 6.1.18 Prove that

1+ t 1 ∞ − = ∑ [ Pn ( x ) + Pn+1 ( x )]t n. t 1 − 2 xt + t 2 t n= 0 [JNTU 2005S (Set 4)]

Solution Since ∞

(1 − 2 xt + t 2 ) − 2 = ∑ Pn ( x )t n 1

(6.67)

n= 0

∴

1+ t 1 1 1 1 1 − = (1 − 2 xt + t 2 ) − 2 + (1 − 2 xt + t 2 ) − 2 − 2 t t t t 1 − 2 xt + t ∞ 1 ∞ 1 = ∑ Pn ( x )t n + ∑ Pn ( x )t n − by Eq. (6.67) t n= 0 t n= 0 ∞ ⎞ ∞ 1⎛ 1 1 1 ∞ = ⎜ P0 ( x ) + ∑ Pnt n ⎟ + ∑ Pn ( x )t n − = + ∑ Pn ( x )t n ⎠ n= 0 t⎝ t t t n=1 n =1 ∞ 1 + ∑ Pn ( x )t n − ∵ P0 ( x ) = 1 t n= 0 ∞

∞

∞

∞

n =1

n= 0

n= 0

n= 0

= ∑ Pn ( x )t n −1 + ∑ Pn ( x )t n = ∑ Pn+1 ( x )t n + ∑ Pn ( x )t n replacing n by ( n + 1) in the first summation ∞

= ∑ [ Pn ( x ) + Pn+1 ( x )]t n n= 0

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6-28

Differential Equations

Example 6.1.19 Expand f (x) in a series of Legendre polynomials if ⎧ x, 0 < x < 1 f ( x) = ⎨ ⎩0, −1 < x < 0. Solution Let f ( x ) = C0 P0 ( x ) + C1 P1 ( x ) +

+ Cn Pn ( x ) +

. Then

⎛ 2n + 1⎞ 1 ⎛ 2n + 1⎞ Cn = ⎜ f ( x ) Pn ( x )dx = ⎜ ⎟ ∫ ⎝ 2 ⎠ −1 ⎝ 2 ⎟⎠ ⎡ 0 0 ⋅ P ( x )dx + 1 xP ( x )dx ⎤ = ⎛ 2n + 1⎞ 1 xP ( x )dx n ∫0 n ⎣⎢ ∫−1 ⎦⎥ ⎜⎝ 2 ⎟⎠ ∫0 n 1 1 1 n = 0, C0 = ∫ xdx = 0 2 4 3 1 2 1 n = 1, C1 = ∫ x dx = 0 2 2 5 1x 5 n = 2, C2 = ∫ (3x 2 − 1)dx = and so on 0 2 2 16 EXERCISE 6.1 1. Show that (a) Pn′ (1) =

1 n n( n + 1); (b) Pn′ ( −1) = ( −1) n −1 ( n + 1). 2 2 [Hint: Use generating functions or (1 − x 2 ) Pn′′ − 2 xPn′ + n( n + 1) Pn = 0]

2. Show that

⎧0 n ≠ 0 Pn ( x )dx = ⎨ . −1 ⎩2 n = 0

∫

1

[Hint: Use orthogonal property.] 1 [8 P ( x ) + 28 P3 ( x ) + 27 P1 ( x )]. 63 5 ( −1) n (2n)! 4. Show that (a) P2 n (0) = 2 n ; (b) P2 n+1 (0) = 0. 2 ( n!) n [JNTU 2006 (Set 2)] n 1 2 t 5. Show that ∫ (1 − 2 xt + t 2 ) −1 2 Pn ( x )dx = . −1 2t + 1 3. Show that x 5 =

6. Show that

Ch06.indd 28

∫

π

0

Pn (cos φ ) cos nφ dφ =

1 ⋅ 3 ⋅ 5 (2n − 1) . 2 ⋅ 4 ⋅ 6 2n

12/13/2011 11:11:57 AM

Integration in Series

7. Show that ( Pn+1 − Pn −1 )

(2 n +1)

6-29

= ∫ Pn dx + c.

2n . 4 n2 − 1 [Hint: Use recurrence relation RR1 and orthogonal property.]

8. Show that

∫

1

9. Show that

∫

1

−1

−1

xPn ( x ) Pn −1 ( x )dx =

x 2 Pn+1 ( x ) Pn −1 ( x )dx =

2n( n + 1) . (2n − 1)(2n + 1)(2n + 3)

10. Show that 2 P3 ( x ) + 3P1 ( x ) = 5 x 3 .

[Hint: Use RR1] [JNTU 2003S (Set 4)] [JNTU 2006 (Set 3)]

6.2 BESSEL FUNCTIONS 6.2.1 Introduction Bessel functions are solutions of Bessel’s differential equation which arises in the solution of Laplace’s equation in cylindrical coordinates. It occurs in many boundary value problems arising in electrical fields, mechanical vibrations and heat conduction.

6.2.2

Bessel Functions

The second order linear differential equation x 2 y ′′ + xy ′ + ( x 2 − p 2 ) y = 0 or

1 d ⎛ dy ⎞ ⎛ p2 ⎞ x + 1 − ⎜ ⎟ ⎜ ⎟ y=0 x dx ⎝ dx ⎠ ⎝ x2 ⎠

(6.68)

is called Bessel’s 4 equation of order p. Its particular solutions are called Bessel functions of order p. p is a real (or complex) constant. Here we assume p to be real non-negative number. As mentioned earlier, Eq. (6.68) can be solved by the method of Frobenius. Writing Eq. (6.68) in the standard form we have y ′′ +

1 p2 ⎞ ⎛ y ′ + ⎜1 − 2 ⎟ y = 0 ⎝ x x ⎠

(6.69)

4

Named after the German mathematician and astronomer Friedrich Wilhelm Bessel (1784–1846) whose paper on Bessel functions dated 1824 appeared in 1826. He started out as an apprentice of a trade company studying astronomy on his own in his spare time, later became an assistant at a small private observatory and finally became director of new Konigsberg observatory.

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6-30

Differential Equations

We assume a solution of Eq. (6.69) in the form ∞

y( x ) = ∑ ar x m+ r = x m ( a0 + a1 x + r=0

+ ar x r +

)

(6.70)

Differentiating Eq. (6.70) w.r.t. x twice, we get ∞

y ′( x ) = ∑ ( m + r )ar x m+ r −1

(6.71)

r=0 ∞

y ′′( x ) = ∑ ( m + r )( m + r − 1)ar x m+ r − 2

(6.72)

r=0

Substituting Eqs. (6.70)–(6.72) into Eq.(6.68) we obtain

∑ [(m + r )( m + r − 1) + (m + r ) − p ]a x 2

r

m+ r

+ ∑ ar x m+ r + 2 = 0.

(6.73)

Equation (6.73) must be identically satisfied. This implies that the coefficient of each power of x must be zero. The lowest power of x is xm obtained for r = 0. Equating to zero the coefficient of xm, we get the indicial equation m2 − p 2 = 0 ⇒ m = p or − p ∵ a0 ≠ 0 Equating to zero the coefficient of [( m + 1)2 − p 2 ]a1 = 0 ⇒ a1 = 0 ∵ ( m + 1)2 ≠ p 2

x m+1 ,

(6.74) we

obtain

Generally equating to zero the coefficient of x m+ r + 2 , we get the coefficient recurrence relation [( m + r + 2)2 − p 2 ]ar + 2 + ar = 0 ar (6.75) ( m − p + r + 2)( m + p + r + 2) Putting r = 1, 3, 5, … we see that the odd coefficients vanish, i.e., a2 r −1 = 0 for all r = 1, 3, 5,…. Putting r = 0, 2, 4, … we get a0 a2 a2 = − , a4 = − ( m − p + 2)( m + p + 2) ( m − p + 4)( m + p + 4) a0 , = ( m − p + 2)( m + p + 2)( m − p + 4) × ( m + p + 4) − a0 a4 etc. a6 = − = ( m − p + 6)( m + p + 6) ( m − p + 2)( m + p + 2)( m − p + 4) ( m + p + 4)( m − p + 6)( m + p + 6) ⇒ ar + 2 = −

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Integration in Series

6-31

A solution of Eq. (6.68) is ⎡ x2 y = a0 x m ⎢1 − 2 2 ⎣ [( m + 2) − p ] x4 + [( m + 2)2 − p 2 ][( m + 4)2 − p 2 ] x6 − + [( m + 2)2 − p 2 ][( m + 4)2 − p 2 ][( m + 6)2 − p2 ]

⎤ ⎥ (6.76) ⎦

We get different types of solutions depending on the values of p.

6.2.3

Bessel Functions of Non-integral Order p: Jp(x) and J-p(x)

Case (i) p is not an integer. In this case, we get two linearly independent solutions for m = p and m = –p. For m = p, we have ⎡ x2 x4 + y1 = a0 x p ⎢1 − ⎣ 2(2 p + 2) 2 ⋅ 4(2 p + 2)(2 p + 4) ⎤ x6 − 2 ⋅ 4 ⋅ 6(2 p + 2)(2 p + 4)(2 p + 6) ⎥⎦ ⎡ x2 x4 = a0 x p ⎢1 + ( −1) + ( −1)2 4 + ( −1)3 + + + 2(1!)( 1) 2 (2!)( 1)( 2) p p p ⎣ ⎤ x6 × 6 + ⎥ 2 (3!)( p + 1)( p + 2)( p + 3) ⎦ ∞ r ( 1) − x 2r = a0 x p ∑ 2 r ( p + r) r = 0 2 ( r !)( p + 1)( p + 2) ∞ ( −1) r Γ ( p + 1) x 2 r = a0 x p ∑ 2 r r = 0 2 Γ ( r + 1)Γ ( p + r + 1) Since a0 is arbitrary we choose it as a0 = 2 p Γ (1p +1) and denote the resulting solution by Jp ( x ). This is called Bessel function of the first kind of order p. Thus ∞

( −1) r ⎛ x⎞ Jp ( x ) = ∑ ⎜⎝ ⎟⎠ r = 0 r ! Γ ( p + r + 1) 2

Ch06.indd 31

p+2r

(6.77)

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6-32

Differential Equations

A second linearly independent solution corresponding to m = − p is ∞

( −1) r ⎛ x⎞ ⎜⎝ ⎟⎠ r = 0 r ! Γ ( − p + r + 1) 2

J − p ( x) = ∑

− p+2r

(6.78)

which is also called Bessel function of the first kind of order –p. Both the series converge for all x as can be seen by D’ Alembert’s ratio test. When p is not an integer, the complete solution of Eq. (6.68) is y( x ) = AJp ( x ) + BJ − p ( x )

(6.79)

where A and B are arbitrary constants.

6.2.4

Bessel Functions of Order Zero and One: J0(x), J1(x)

Case (ii) p = 0 Putting p = 0 in Eq. (6.68) we obtain Bessel’s equation of order zero as xy ′′ + y ′ + xy = 0

or ( xy ′ ) ′ + xy = 0

(6.80)

Its solution obtained from Eq. (6.76) by putting p = 0 and taking a0 = 1 is ⎡ x2 x4 y = x m ⎢1 − + 2 ( m + 2)2 ( m + 4)2 ⎣ ( m + 2) ⎤ x6 − + ⎥ ( m + 2)( m + 4)( m + 6) ⎦

(6.81)

which is the solution of Eq. (6.80) if m = 0. Thus, the first solution of Eq. (6.80) is ∞

2r

( −1) r ⎛ x ⎞ ⎜⎝ ⎟⎠ 2 r = 0 ( r !) 2 x x4 x6 =1− 2 2 + 4 − + 2 (1!) 2 (2!)2 26 (3!)2

J 0 ( x) = ∑

(6.82)

which is called Bessel’s function of the first kind of order zero, and appears to be similar to the cosine function.

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Integration in Series

6-33

For p = 1, we obtain the Bessel function of order 1 ∞

( −1) r x 2 r +1 x x3 = − 2 r +1 r !( r + 1)! 2 231!2! r=0 2 5 x x7 (6.83) + 5 − 7 + 2 ⋅ 2!3! 2 ⋅ 3!4! This looks similar to the sine function. The height of the waves decreases with increasing x. Dividing Bessel’s Eq. (6.68) by x2 we can put it in the standard form J1 ( x ) = ∑

(

)

y ′′ + 1x y + 1 − nx2 y = 0 where p has been replaced by n. The term 2

n2 x2

is 0 for n = 0. Further, this term as well as the term 1x y′ are small in absolute value for large x so that the Bessel’s equation comes closer to y ′′ + y = 0 whose solutions are sine and cosine functions. Also, 1x y acts as a damping term which is partly responsible for the decrease in height (see Fig. 6.2). For large x, we can derive the result that J n ( x) = 1 0.5 0

2 2n + 1 ⎞ ⎛ π⎟ . cos ⎜ x − ⎝ ⎠ πx 4

J0 J1 5

10

x

Figure 6.2 Bessel functions of the first kind

6.2.5

Bessel Function of Second Kind of Order Zero Y0(x)

Differentiating Eq. (6.81) partially w.r.t. m, we get ⎡ x2 x4 ∂y = x m log x ⎢1 − + 2 ( m + 2)2 ( m + 4)2 ∂m ⎣ ( m + 2) ⎤ x6 − + … ⎥ ( m + 2)2 ( m + 4)2 ( m + 6)2 ⎦ 2 4 ⎡ 2 2 ⎫ x x ⎧ 2 + xm ⎢ ⋅ − ×⎨ + ⎬+ 2 2 2 ⎣ ( m + 2) m + 2 ( m + 2) ( m + 4) ⎩ m + 2 m + 4 ⎭

Ch06.indd 33

⎤ ⎥ ⎦

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6-34

Differential Equations

A second independent solution of Eq. (6.80) denoted by Y0 ( x ) is given by ∂∂my

( )

m= 0

4 ⎡⎛ x ⎞ 2 1 ⎛ 1⎞ ⎛ x ⎞ + Y0 ( x ) = J 0 ( x ) log x ⎢⎜ ⎟ − 1 ⎜ ⎟⎜ ⎟ (2!)2 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎢⎣⎝ 2 ⎠ 6 ⎤ 1 ⎛ 1 1⎞ ⎛ x ⎞ + 1+ + ⎟ ⎜ ⎟ − ⎥ 2 ⎜ (3!) ⎝ 2 3 ⎠ ⎝ 2 ⎠ ⎥⎦ ∞

( −1) r +1 ⎛ 1 1 × ⎜1 + + + 2 ⎝ 2 3 r =1 ( r !)

= J 0 ( x ) log x + ∑

1⎞ ⎛ x ⎞ + ⎟⎜ ⎟ r⎠ ⎝ 2⎠

2r

(6.84)

Y0 ( x ) is called the Bessel function of the second kind of order zero or Neumann function. So, the complete solution in Bessel Eq. (6.80) of order zero is y( x ) = AJ 0 ( x ) + BY0 ( x )

(6.85)

where A and B are arbitrary constants.

6.2.6

Bessel Functions of Integral Order: Linear Dependence of Jn(x) and J-n(x)

Case (iii) p = n (an integer): To prove that J − n ( x ) = ( −1) n J n ( x ). Taking p = –n, a positive integer so that –n is a negative integer, we obtain from Eq. (6.78) ∞

− n+ 2 r

( −1) r ⎛ x⎞ ⎜⎝ ⎟⎠ 2 r = 0 r ! Γ ( − n + r + 1) 2( r − n ) + n ∞ ( −1) r − n ⎛ x⎞ n = ( −1) ∑ ⎜⎝ ⎟⎠ 2 ( r − n ) = 0 ( r − n + n)! Γ ( r − n + 1) ∵ Γ ( r − n + 1) = ∞ for r < n 2s+ n ∞ ( −1) s ⎛ x⎞ = ( −1) n ∑ ⎜⎝ ⎟⎠ 2 s = 0 ( s + n)! Γ ( s + 1) where s = r − n ≥ 0 = ( −1) n J n ( x ). (6.86)

J − n ( x) = ∑

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Integration in Series

6-35

This proves that when p is an integer Bessel functions Jp ( x ) and J− p ( x ) are linearly dependent.

6.2.7

Bessel Functions of the Second Kind of Order n: Yn(x): Determination of Second Solution Yn(x) by the Method of Variation of Parameters

Since J n ( x ) and J − n ( x ) are linearly dependent when n is an integer, the second solution of Eq. (6.68) is obtained by the following method. Let y = uv, where v = J n ( x ), be a solution of Eq. (6.68). Then y ′ = u ′v + uv ′,

y ′′ = u ′′v + 2u ′v ′ + uv ′′

Substituting in Eq. (6.68), we get x 2 [u ′′v + 2u ′v ′ + uv ′′ ] + x[u ′v + uv ′ ] + ( x 2 − n2 )uv = 0 ⇒ x 2 [u ′′v + 2u ′v ′ ] + xu ′v = 0. The coefficient of u is zero since v = J n ( x ) is a solution of Eq. (6.68). u ′′ v′ 1 ⇒ + 2 + = 0, on dividing each term by x 2 u ′v u′ v x ⇒ (log u ′ ) ′ + 2(log v ) ′ + (log x ) ′ = 0 ⇒ log( xu ′ ⋅ v 2 ) = log B, on integration w.r.t. x dx ⇒ u( x ) = B ∫ 2 + A, on dropping logs, after dividing by xv2 xJ n ( x ) and integrating. Hence the complete solution of Eq. (6.68) when p = n, an integer, is y( x ) = AJ n ( x ) + BYn ( x ) where Yn ( x ) = J n ( x ) ∫

6.2.8

dx , and A and B are arbitrary constants. xJ n2 ( x )

Generating Functions for Bessel Functions

Let < f n > where n ∈

be a sequence of functions fn. A function

w(t, x) such that w (t , x ) =

∑

∞

n = −∞

f n ( x ) is called a generating function of

the functions f n as already given at Section 6.1.6 (p. 6–12).

Ch06.indd 35

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6-36

Differential Equations

Generating function for Bessel functions Jn(x) of integral order n Example 6.2.1 Prove that e 2 (t − t ) = x

1

∞

∑t

n

n = −∞

Solution

(6.87)

J n ( x ).

⎡ 1 ⎛ x⎞ 1 1 ⎛ x⎞ 1 ⎛ x⎞ x x e (t − t ) = e 2 t ⋅ e − 2 t = ⎢1 + ⎜ ⎟ t + ⎜ ⎟ t 2 + + ⎜ ⎟ t n + 2! ⎝ 2 ⎠ n! ⎝ 2 ⎠ ⎢⎣ 1! ⎝ 2 ⎠ 2 n ⎡ 1 ⎛ x⎞ 1 1 ⎛ x⎞ 1 ⎤ 1 ⎛ x⎞ 1 × ⎢1 − ⎜ ⎟ + ⎜ ⎟ 2 − + ( −1) n ⎜ ⎟ n + ⎥ n! ⎝ 2 ⎠ t ⎣⎢ 1! ⎝ 2 ⎠ t 2! ⎝ 2 ⎠ t ⎦⎥ n

2

x 2

⎤ ⎥ ⎥⎦

In the product on the RHS Coefficient of 2

4

6

1 ⎛ x⎞ 1 ⎛ x⎞ 1 ⎛ x⎞ t =1− + − ⋅⎜ ⎟ − ⎟ ⎜ ⎟ 2 ⎜ 2 (1!) ⎝ 2 ⎠ (2!) ⎝ 2 ⎠ (3!)2 ⎝ 2 ⎠ 2n n! ⎛ x ⎞ + ( −1) ⎜ ⎟ + ( n!)2 ⎝ 2 ⎠ 2n ( −1) r ⎛ x ⎞ =∑ ⎜ ⎟ = J 0 ( x) ( r !)2 ⎝ 2 ⎠ 0

(6.88)

Coefficient of n

n+ 2

1 ⎛ x⎞ 1 1 ⎛ x⎞ ⎛ x⎞ ⎜⎝ ⎟⎠ − ⎜⎝ ⎟⎠ + ⎜ ⎟ n! 2 ( n + 1)!1! 2 ( n + 2)!2! ⎝ 2 ⎠ n+ 2 r ( −1) r ⎛ x ⎞ =∑ = J n ( x) ⎜ ⎟ r !( n + r )! ⎝ 2 ⎠

tn =

n+ 4

− (6.89)

Coefficient of t

−n

n+ 2 n+ 4 ⎡ 1 ⎛ x⎞ n 1 1 ⎛ x⎞ ⎛ x⎞ = ( −1) ⎢ ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ − ( n + 1)!1! ⎝ 2 ⎠ ( n + 2)!2! ⎝ 2 ⎠ ⎢⎣ n! ⎝ 2 ⎠ n+ 2 r ( −1) r ⎛ x ⎞ = ( −1) n ∑ = ( −1) n J n ( x ) = J − n ( x ) ⎜ ⎟ r !( n + r )! ⎝ 2 ⎠ n

⎤ ⎥ ⎥⎦

(6.90)

The result follows from Eqs. (6.88), (6.89) and (6.90).

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Integration in Series

6.2.9

6-37

Recurrence Relations of Bessel Functions

Example 6.2.2 The following recurrence relations connect Bessel functions of different orders and are very useful in solving problems involving Bessel functions. These relations are true for general p but we prove them taking p = n. d n ( x J n ( x )) = x n J n −1 ( x ) dx d −n ( x J n ( x )) = − x − n J n +1 ( x ) RR2: dx n RR3: J n′ ( x ) + J n ( x ) = J n −1 ( x ) x n RR4: J n′ ( x ) − J n ( x ) = J n +1 ( x ) x RR1:

RR5: 2 J n′ ( x ) = J n −1 ( x ) − J n+1 ( x ) RR6:

2n J ( x ) = J n −1 ( x ) + J n+1 ( x ) x n

Solution To prove RR1 d n ( x J n ( x )) = x n J n −1 ( x ) dx ∞

( −1) r ⎛ x⎞ ⎜⎝ ⎟⎠ r = 0 r ! Γ ( n + r + 1) 2

We know that J n ( x ) = ∑

n+ 2 r

∞

( −1) r x 2 n+ 2 r n+ 2 r r !( n + r )Γ ( n + r ) r=0 2 ∞ d n ( −1) r x 2 n + 2 r −1 ⇒ [ x J n ( x )] = ∑ n + 2 r −1 = x n J n −1 ( x ) dx r !Γ ( n + r ) r=0 2 ⇒

x n J n ( x) = ∑

To prove RR2 d −n ( x J n ( x )) = − x − n J n+1 ( x ) dx

Ch06.indd 37

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6-38

Differential Equations ∞

( −1) r ⎛ x⎞ ⎜⎝ ⎟⎠ r = 0 r ! Γ ( n + r + 1) 2

J n ( x) = ∑

We know that

n+ 2 r

∞

( −1) r x 2 r n+ 2 r r !Γ ( n + r + 1) r=0 2 ∞ d −n ( −1) r −1 x 2 r −1 ⇒ [ x ⋅ J n ( x )] = − ∑ n + 2 r −1 dx ( r − 1)!Γ ( n + 1 + r ) r =1 2 n +1+ 2 s ∞ ( −1) s ⎛ x⎞ = − x−n ∑ ×⎜ ⎟ ⎝ 2⎠ s = 0 s ! Γ ( n + 1 + s + 1) −n = − x J n+1 ( x ), replacing r − 1 by s ⇒ x − n J n ( x) = ∑

To prove RR3 J n′ ( x ) +

n J ( x ) = J n −1 ( x ) x n

d n ( x J n ( x )) = x n J n −1 ( x ) dx ⇒ x n J n′ ( x ) + nx n −1 J n ( x ) = x n J n −1 ( x ) n ⇒ J n′ ( x ) + J n ( x ) = J n −1 ( x ) x

RR1 ⇒

on dividing by x n . To prove RR4 J n′ ( x ) −

n J ( x ) = − J n+1 ( x ) x n

d −n ( x J n ( x )) = − x − n J n +1 ( x ) dx ⇒ x − n J n′ ( x ) − nx − n −1 J n ( x ) = − x − n J n +1 ( x ) n ⇒ J n′ ( x ) − J n ( x ) = − J n +1 ( x ) x

RR2 ⇒

on dividing by x − n . To prove RR5 2 J n′ ( x ) = J n −1 ( x ) − J n+1 ( x )

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Integration in Series

6-39

n J ( x ) = J n −1 ( x ) x n n and RR4 ⇒ J n′ ( x ) − J n ( x ) = − J n+1 ( x ) x We get 2 J n′ ( x ) = J n −1 ( x ) − J n+1 ( x ) Adding RR3 ⇒ J n′ ( x ) +

To prove RR6 n 2 J n ( x ) = J n −1 ( x ) + J n+1 ( x ) x n J ( x ) = J n −1 ( x ) x n n RR4 ⇒ J n′ ( x ) − J n ( x ) = − J n+1 ( x ) x Subtracting, we get We have RR3 ⇒ J n′ ( x ) +

n 2 J n ( x ) = J n −1 ( x ) + J n+1 ( x ) x

6.2.10

Bessel’s Functions of Half-integral Order

Bessel function J p of orders p = n ± 12 ( n = 0,1, 2, 3,…) are elementary functions. They can be expressed in terms of sines and cosines and powers of x. Example 6.2.3 Prove that 2 sin x; πx 2 (ii) J −1 2 ( x ) = cos x; πx 2 (iii) J122 ( x ) + J −21 2 ( x ) = . πx

(i) J1 2 ( x ) =

Solution (i) We know that J p ( x) =

Ch06.indd 39

xp 2 p Γ ( p + 1)

⎡ x2 x4 ⎢1 − 2 ⋅ 2( p + 1) + 2 ⋅ 4 ⋅ 22 ( p + 1)( p + 2) − ⎣

⎤ ⎥ ⎦

12/13/2011 11:12:01 AM

6-40

Differential Equations

Putting p =

1 2

⎡ x1 2 x2 x4 1 − + − ⎢ 21 2 Γ ( 12 + 1) ⎣ 2 ⋅ 2 ( 12 + 1) 2 ⋅ 4 ⋅ 22 ( 12 + 1)( 12 + 2) x1 2 x3 x5 ⎡ ⎤ = − + − ⎥ x ⎢ 1 12 2 ⋅ 3 2 ⋅ 4 ⋅ 3⋅ 5 2 x⋅ 2 π ⎣ ⎦ 3 5 2 ⎡ 2 x x ⎤ = − ⎥,= sin x x− + ⎢ πx ⎣ πx 3! 5! ⎦

J 1 ( x) = 2

⎤ ⎥ ⎦

on multiplying and dividing by x, and noting ⎛ 1 ⎞ 1 ⎛ 1⎞ 1 π Γ ⎜ + 1⎟ = Γ ⎜ ⎟ = ⎝ 2 ⎠ 2 ⎝ 2⎠ 2 (ii) We know that J p ( x) =

⎡ xp x2 x4 1 − + − 2 p Γ ( p + 1) ⎢⎣ 2 ⋅ 2( p + 1) 2 ⋅ 4 ⋅ 22 ( p + 1)( p + 2)

Putting p = − J − 1 ( x) = 2

+ =

⎤ ⎥ ⎦

1 2

⎡ x −1 2 x2 1 − ⎢ 2 −1 2 Γ ( − 12 + 1) ⎣ 2 ⋅ 2 ( − 12 + 1)

⎤ ⎥ ⎦ 2 ⎤ ⎥⎦ = π x cos x

x4 − 2 ⋅ 4 ⋅ 22 ( − 12 + 1)( − 12 + 2) 2 πx

⎡ x2 x4 ⎢⎣1 − 2! + 4! −

⎛ 1⎞ ∵Γ ⎜ ⎟ = π ⎝ 2⎠

(iii) Squaring and adding the above results we have 2 2 J 12 ( x ) + J −2 1 ( x ) = (sin 2 x + cos 2 x ) = 2 2 πx πx Example 6.2.4 Prove that 2 ⎛ sin x ⎞ − cos x ⎟ ⎜ ⎠ πx ⎝ x 2 ⎛ cos x ⎞ (ii) J −3 2 ( x ) = − ⎜⎝ sin x + ⎟. πx x ⎠ (i) J 3 2 ( x ) =

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Integration in Series

6-41

Solution 2p J ( x ) = J p −1 ( x ) + J p +1 ( x ), where p is taken in place of n. x p 1 Taking p = we have 2

(i) RR6 :

J 3 ( x) = 2

(ii) Taking p = − RR6 :

1 2 ⎛ sin x ⎞ − cos x ⎟ J 1 ( x) − J − 1 ( x) = ⎜⎝ 2 2 ⎠ πx x x 1 in 2

2p J ( x ) = J p −1 ( x ) + J p +1 ( x ), we have x p 2 ( − 12 ) x

or

6.2.11

J − 1 ( x) = J − 3 ( x) + J 1 ( x) 2

2

2

1 1 2 2 J − 3 ( x) = − J − 1 ( x) − J 1 ( x) = − ⋅ cos x − sin x 2 2 πx x 2 x πx 2 ⎛ cos x ⎞ =− + sin x ⎟ ⎜⎝ ⎠ πx x

Differential Equation Reducible to Bessel’s Equation

The differential equation d2 y dy x 2 2 + x + ( λ 2 x 2 − p2 ) y = 0 dx dx where λ is a parameter, can be written as d2 y dy ( λ x )2 + (λ x) + [( λ x )2 − p 2 ] y = 0 2 d (λ x) d (λ x) d2 y dy +t + (t 2 − p 2 ) y = 0 (t = λ x ) 2 dt dt which is Bessel’s equation. When p is a non-integer the general solution is y = C1 J p (t ) + C2 J − p (t ) = C1 J p ( λ x ) + C2 J − p ( λ x ) or

Ch06.indd 41

t2

12/13/2011 11:12:02 AM

6-42

Differential Equations

and when p = n is an integer the general solution is y = C1 J n (t ) + C2Yn (t ) = C1 J n ( λ x ) + C2Yn ( λ x )

6.2.12

Orthogonality

Definition of orthogonality of functions A set of functions f1 , f 2 ,… defined on some interval I = [a, b,] is said to be orthogonal on I with respect to a weight function w ( x ) > 0 if

∫

b

a

w ( x ) f m ( x ) f n ( x )dx = 0 for m ≠ n

The norm f m of f m is defined by fm =

∫

b

a

w ( x ) f m2 ( x )dx

The functions are called orthonormal of I if they are orthogonal on I and all have norm equal to unity. For ‘orthogonal w.r.t. w(x) = 1’ we simply say orthogonal on I. Thus, functions f1 , f 2 ,… are orthogonal on some interval I if

∫

b

a

f m ( x ) f n ( x )dx = 0 for m ≠ n.

The norm f m of f m is then simply defined by fm =

∫

b

a

f m2 ( x )dx

and the functions are called orthonormal on a ≤ x ≤ b if they are orthogonal there and all have norm equal to unity. Orthogonal set, orthonormal set Example 6.2.5 Prove that the functions f m ( x ) = sin mx ( m = 1, 2, 3,…) form an orthogonal set on −π ≤ x ≤ π . Solution Here a = −π , b = π , f m ( x ) = sin mx and we have

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Integration in Series

∫

b

a

π

f m ( x ) f n ( x )dx = ∫ sin m sin nxdx = −π

6-43

1 π cos( m − n) xdx 2 ∫− π

1 π cos( m + n) xdx 2 ∫− π π π 1 sin( m − n) x 1 sin( m + n) x = ⋅ − ⋅ =0 m − n −π 2 m + n −π 2 −

since the functions vanish at both the limits. Also, f m

2

π

1 π 1 (1 − cos 2mx )dx = ⋅ 2π − 0 = π ∫ π − 2 2

= ∫ sin 2 mxdx = −π

∴ The norm = π ; hence the orthonormal set is sin x sin 2 x sin 3 x , , ,

π

π

π

Orthogonality of Bessel functions Example 6.2.6 Prove that, for each fixed non-negative integer n,

∫

a

0

if α ≠ β ⎧0, ⎪ 2 xJ n (α x ) J n (β x )dx = ⎨ a 2 ⎪⎩ 2 J n+1 ( aα ), if α = β

where a and b are roots of J n ( ax ) = 0. Solution Let u( x ) = J n (α x ) and v( x ) = J n (β x ) be the solutions of the equations x 2 u ′′ + xu ′ + (α 2 x 2 − n2 ) u = 0 (6.91) x 2 v ′′ + xv ′ + (β 2 x 2 − n2 ) v = 0

(6.92)

respectively. Hence u ′( x ) = α J n′ (α x ), v ′( x ) = β J n′ (β x ). Now multiplying Eq. (6.91) by

v x

and Eq. (6.92) by

u x

and subtracting.

x(u ′′v − uv ′′ ) + (u ′v − uv ′ ) + (α 2 − β 2 ) xuv = 0 ⇒

Ch06.indd 43

d [ x(u ′v − uv ′ )] = (β 2 − α 2 ) xuv dx

(6.93)

12/13/2011 11:12:03 AM

6-44

Differential Equations

Integrating both sides of Eq. (6.93) w.r.t. x from x = 0 to a a

(β 2 − α 2 ) ∫ xuvdx = [ x(u ′v − uv ′ )]0a 0

= a[u ′( a)v( a) − u( a)v ′( a)]

(6.94)

This implies that a

(β 2 − α 2 ) ∫ xJ n (α x ) J n (β x )dx = a[α J n′ (α a) J n (β a) 0

−β J n′ (β a) J n (α a)]

a

⇒ ∫ xJ n (α x ) J n (β x )dx = 0

a [α J n′ (α a) J n (β a) β − α2 − β J n′ (β a) J n (α a)] 2

(6.95)

Case (i) α ≠ β That is, a, b are two distinct roots of J n ( ax ) so that J n ( aα ) = J n ( aβ ) = 0. Therefore, we obtain from Eq. (6.95)

∫

a

0

xJ n (α x ) J n (β x )dx = 0

which is the orthogonality relation for Bessel functions. Case (ii) α = β In this case, the RHS of Eq. (6.95) assumes the indeterminate form 00 . Hence we apply L’ Hospital’s Rule by differentiating w.r.t. b and evaluating the limits. a ⎛ a ⎞ lim ∫ xJ n (α x ) J n (β x )dx = lim ⎜ 2 β →α ⎝ β − α 2 ⎟ ⎠ [α J n′ ( aα ) J n ( aβ ) − 0], ∵ J n ( aα ) = 0 a = lim [α J n′ ( aα ).aJ n′ ( aβ )] β →α 2β a2 a2 2 = [ J n′ ( aα )]2 = J ( aα ). 2 2 n+1 By recurrence relation RR4: n J n′ ( x ) = J n ( x ) − J n+1 ( x ) x n J ( aα ) − J n+1 ( aα ) = 0 − J n+1 ( aα ) ⇒ J n′ ( aα ) = aα n = − J n+1 ( aα ), (∵ J n ( aα ) = 0)

β →α 0

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Integration in Series

6-45

If we put x = aα in the recurrence relation RR6, we have J n −1 ( aα ) + J n+1 ( aα ) =

2n J ( aα ) = 0 aα n

Thus for α = β

∫

a

0

xJ n (α x ) J n (β x )dx =

a2 a2 J n+1 ( aα ) = J ( aα ) 2 2 n −1

6.2.13 Integrals of Bessel Functions

∫x

Example 6.2.7 Prove that

−p

J p +1 ( x )dx = − x − p J p ( x ) + c.

Solution Recurrence relation RR1 is d p ( x J p ( x )) = x p J p −1 ( x ) dx Integrating both sides we have

∫x

p

J p −1 ( x )dx = x p J p ( x ) + c

For p = 1, we get

∫ xJ

0

( x )dx = xJ1 ( x ) + c

Recurrence relation RR2 is d −p ( x J p ( x )) = − x − p J p +1 ( x ) dx Integrating both sides, we have

∫x

−p

J p +1 ( x )dx = − x − p J p ( x ) + c

For p = 0 we get

∫ J ( x)dx = − J 1

0

( x) + c

In general, ∫ x m J n ( x )dx for m and n integer with m + n ≥ 0 can be integrated by parts completely if m + n is odd. But when m + n is even the integral depends on the residual integral ∫ J 0 ( x )dx which has been tabulated. 1 J p′ ( x ) = [ J p −1 ( x ) − J p +1 ( x )] Integrating, 2

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6-46

Differential Equations

We get 2 J p ( x ) = ∫ J p −1 ( x )dx − ∫ J p +1 ( x )dx or

∫J

6.2.14

p +1

( x )dx = ∫ J p −1 ( x )dx − 2 J p ( x )

Expansion of Sine and Cosine in Terms of Bessel Functions

Example 6.2.8 Show that (i) cos x = J 0 − 2 J 2 + 2 J 4 − (ii) sin x = 2( J1 − J 3 + J 5 −

).

Solution We know that ∞

e 2 (t − t ) = x

1

∑t

n = −∞

n

−1

∞

−∞

1

J n ( x) = ∑ t n J n ( x) + J 0 ( x) + ∑ t n J n ( x) ∞

= J 0 ( x ) + ∑ [t − n J − n ( x ) + t n J n ( x )] n =1

1⎞ ⎛ 1⎞ ⎛ = J 0 ( x ) + ⎜ t − ⎟ J1 ( x ) + ⎜ t 2 + 2 ⎟ J 2 ( x ) + ⎝ t⎠ ⎝ t ⎠ ∞

= J 0 ( x ) + ∑ (t n + ( −1) n t − n ) J n ( x ) n =1

Put t = e iθ e x ( e − e ) 2 = J 0 + (e iθ − e − iθ ) J1 + (e 2iθ + e −2iθ ) J 2 + ⇒ e ix sin θ = J 0 + (2i sin θ ) J1 + (2cos 2θ ) J 2 + iθ

− iθ

Separating the real and imaginary parts we have cos( x sin θ ) = J 0 + 2cos 2θ J 2 + 2cos 4θ J 4 + sin( x sin θ ) = 2sin θ J1 + 2sin 3θ J 3 + 2sin 5θ J 5 + Putting θ = π 2, we get cos x = J 0 − 2 J 2 + 2 J 4 + sin x = 2( J1 − J 3 + J 5 − )

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Integration in Series

6-47

Example 6.2.9 Find series expansion for J 0 ( x ) and J1 ( x ). Solution We know that ∞

( −1) r ⎛ x ⎞ ⎜⎝ ⎟⎠ r = 0 r !( n + r )! 2

J n ( x) = ∑

n+ 2 r

Putting n = 0 we get 2r

∞

( −1) r ⎛ x ⎞ 2 ⎜ ⎝ 2 ⎟⎠ r = 0 ( r !) 2 4 6 1 ⎛ x⎞ 1 ⎛ x⎞ 1 ⎛ x⎞ =1− ⎜ ⎟ + ⎜ ⎟ − ⎜ ⎟ + (1!)2 ⎝ 2 ⎠ (2!)2 ⎝ 2 ⎠ (3!)2 ⎝ 2 ⎠

J 0 ( x) = ∑

Putting n = 1 we get 2 r +1

∞

( −1) r ⎛ x ⎞ ⎜⎝ ⎟⎠ r = 0 r !( r + 1)! 2 2 4 6 x⎡ 1 ⎛ x⎞ 1 ⎛ x⎞ 1 ⎛ x⎞ = ⎢1 − + − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + 2 ⎣⎢ 1!2! ⎝ 2 ⎠ 2!3! ⎝ 2 ⎠ 3!4! ⎝ 2 ⎠

J1 ( x ) = ∑

⎤ ⎥ ⎦⎥

Example 6.2.10 Express J 5 ( x ) in terms of J 0 ( x ) and J1 ( x ). Solution 2n J ( x ) = J n −1 ( x ) + J n+1 ( x ) x n 2n ⇒ J n+1 ( x ) = J ( x ) − J n −1 ( x ) x n

RR6 :

(6.96)

Putting n = 1, 2, 3, 4 we get 2 J − J0 x 1 4 J 3 = J 2 − J1 x 6 J4 = J3 − J2 x 8 J5 = J4 − J3 x J2 =

Ch06.indd 47

(6.97) (6.98) (6.99) (6.100)

12/13/2011 11:12:05 AM

6-48

Differential Equations

where the argument x has been omitted for convenience. Substituting from Eq. (6.97), Eq. (6.98) becomes J3 =

4⎛2 4 ⎞ ⎛ 8 ⎞ ⎜⎝ J1 − J 0 ⎟⎠ − J1 = ⎜⎝ 2 − 1⎟⎠ J1 − J 0 x x x x

(6.101)

Substituting Eqs. (6.101) and (6.97) into Eq. (6.99) 6 ⎡⎛ 8 4 ⎤ ⎡2 ⎞ ⎤ ⎜⎝ 2 − 1⎟⎠ J1 − J 0 ⎥ − ⎢ J1 − J 0 ⎥ ⎢ x⎣ x x ⎦ ⎣x ⎦ ⎛ 48 8 ⎞ ⎛ 24 ⎞ = ⎜ 3 − ⎟ J1 + ⎜1 − 2 ⎟ J 0 ⎝x ⎝ x⎠ x ⎠

J4 =

(6.102)

Substituting Eqs. (6.102) and (6.101) into Eq. (6.100), 8 ⎡⎛ 48 8 ⎞ 4 ⎤ ⎛ 24 ⎞ ⎤ ⎡⎛ 8 ⎞ ⎜⎝ 3 − ⎟⎠ J1 + ⎜⎝1 − 2 ⎟⎠ J 0 ⎥ − ⎢⎜⎝ 2 − 1⎟⎠ J1 − J 0 ⎥ ⎢ x⎣ x x x x ⎦ ⎦ ⎣ x ⎛ 384 72 ⎞ ⎛ 12 192 ⎞ ⇒ J 5 = ⎜ 4 − 2 − 1⎟ J1 + ⎜ − 3 ⎟ J 0 ⎝ x ⎠ ⎝ x x x ⎠ J5 =

Example 6.2.11 Express J 5 2 ( x ) in terms of sine and cosine functions. Solution RR6: Put n =

2n J ( x ) = J n −1 ( x ) + J n+1 ( x ) x n

3 3 we get J 3 2 ( x ) = J 1 ( x ) + J 5 2 ( x ) 2 2 x 3 J ( x ) − J1 2 ( x ) x 32 3 2 ⎛ sin x 2 ⎞ = ⋅ − cos x ⎟ − sin x ⎜⎝ ⎠ πx x πx x ⎤ 2 ⎡⎛ 3 − x 2 ⎞ 3 = ⎜⎝ 2 ⎟⎠ sin x − cos x ⎥ ⎢ πx ⎣ x x ⎦ 2 2 sin x; J − 1 ( x ) = cos x J 1 ( x) = 2 2 πx πx 2 ⎛ sin x ⎞ − cos x⎟ . J 3 2 ( x) = ⎜⎝ ⎠ πx x

⇒ J 5 2 ( x) =

Ch06.indd 48

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Integration in Series

6-49

Example 6.2.12 Prove that (i) J − n ( x ) = ( −1) n J n ( x ) (ii) J n ( − x ) = ( −1) n J n ( x ) (iii) J n ( x ) is an even or odd function of x according as n is even or odd. Solution (i) We know that for any non-negative real p. p+2r

∞

( −1) r ⎛ x⎞ J p ( x) = ∑ ⎜⎝ ⎟⎠ r = 0 r ! Γ ( p + r + 1) 2 Let p = − n where n is a positive integer. Then ∞

( −1) r ⎛ x⎞ J − n ( x) = ∑ ⎜⎝ ⎟⎠ r = 0 r ! Γ ( − n + r + 1) 2

(6.103)

− n+ 2 r

(6.104)

Since lim Γ (t ) = ∞, as t → 0 or a negative integer, each term in the above summation is zero as long as −n + r + 1 ≤ 0

(6.105)

i.e., as long as r ≤ ( n − 1). So, the summation starts with r = n, ∞

− n+ 2 r

( −1) r ⎛ x⎞ ⎜⎝ ⎟⎠ r = n r ! Γ ( − n + r + 1) 2 n+ 2 s ∞ ( −1) s ⎛ x ⎞ n = ( −1) ∑ ⎜⎝ ⎟⎠ s = 0 ( n + s)! s ! 2 where s = r − n or r = n + s = ( −1) n J n ( x ) J − n ( x) = ∑

(6.106)

(ii) Let n be a non-negative integer. Replacing x by –x, we have ∞

n+ 2 r

( −1) r ⎛ −x⎞ J n ( − x) = ∑ ⎜⎝ ⎟⎠ 2 r = 0 r ! Γ ( n + r + 1) n+ 2 r ∞ r ( −1) ⎛ x⎞ = ( −1) n ∑ ⎜⎝ ⎟⎠ r = 0 r ! Γ ( n + r + 1) 2 ∴ ( −1) n+ 2 r = ( −1) n ( − 1)2 r = ( −1) n J n ( − x ) = ( −1) n J n ( x ) ⇒

Ch06.indd 49

(6.107)

12/13/2011 11:12:06 AM

6-50

Differential Equations

Let n be a negative integer and n = − m where m is a positive integer. Then J n ( x ) = J − m ( x ) = ( −1) m J m ( x )

(6.108)

Replacing x by –x, we have J n ( − x ) = ( −1) m J m ( − x ) = ( −1) m ⋅ ( −1) m J m ( x ) by Eq. (6.107) ∵ m > 0 = ( −1) m J − m ( x ) by Eq. (6.108) = ( −1)2 m ( −1) − m J − m ( x ) = ( −1) − m J − m ( x ) ∵ ( −1)2 m = 1 = ( −1) n J n ( x ) ∵ − m = n (iii) We have from (ii) J n ( − x ) = ( −1) n J n ( x ) ⇒ J n ( − x) = J n ( x) and J n ( − x ) = − J n ( x )

if n is even ∵ ( −1) n = 1 if n is odd ∵ ( −1) n = −1.

Thus J n is even or odd according as n is even or odd. Example 6.2.13 Prove that (a)

d 2 n +1 2 ⎞ ⎛n ( J n + J n2+1 ) = 2 ⎜ J n2 − J ⎟ ⎝ dx x x n+1 ⎠

1 J n J n′ + J n+1 J n′+1 = [nJ n2 − ( n + 1) J n+1 ] x

or (b)

d n +1 2 ⎞ ⎛n ( xJ n J n+1 ) = 2 ⎜ J n2 − J ⎟. ⎝x dx x n+1 ⎠

Solution d 2 ( J + J n2+1 ) = 2 J n J n′ + 2 J n+1 J n′+1 dx n n Now, RR4 : J n′ = J n − J n+1 x n n +1 RR3 : J n′ = − J n + J n −1 ⇒ J n′+1 = − J + Jn x x n +1

(a)

(6.109) (6.110) (6.111)

replacing n by ( n + 1). Substituting for J n′ and J n′+1 into Eq. (6.109)

Ch06.indd 50

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Integration in Series

6-51

d 2 ⎛n ⎞ ⎛ n +1 ⎞ ( J n + J n2+1 ) = 2 J n ⎜ J n − J n+1 ⎟ + 2 J n+1 ⎜ − J n+1 + J n ⎟ ⎝ ⎠ ⎝ ⎠ dx x x n +1 2 ⎞ ⎛n = 2 ⎜ J n2 − J ⎟ ⎝x x n+1 ⎠ (b)

d ( xJ n J n+1 ) = J n J n+1 + x( J n+1 J n′ + J n J n′+1 ) dx n RR4: J n′ = J n − J n+1 ; x n +1 RR3: J n′+1 = − J + Jn. x n+1 Substituting in Eq. (6.112), we get

(6.112) (6.113) (6.114)

d ⎡⎛ n ⎤ ⎞ n +1 ( xJ n J n+1 ) = J n J n+1 + x ⎢⎜ J n J n+1 − J n2+1 ⎟ − J n J n+1 + J n2 ⎥ ⎠ dx x ⎣⎝ x ⎦ 2 2 = x( J n − J n+1 ) Example 6.2.14 Prove that J 02 + 2( J12 + J 22 + 1 that J 0 ≤ 1 and J n ≤ for n ≥ 1. 2 Solution We have

) = 1. Hence show

d 2 n +1 2 ⎞ ⎛n ( J n + J n2+1 ) = 2 ⎜ J n2 − J ⎟ ⎝x dx x n+1 ⎠

Putting n = 0,1, 2, 3,… d 2 1 ⎞ ⎛ ( J 0 + J12 ) = 2 ⎜ 0 − J12 ⎟ ∵ lim J n = 0 n→ 0 ⎝ dx x ⎠ d 2 1 2 ⎛ ⎞ ( J + J 22 ) = 2 ⎜ J12 − J 22 ⎟ ⎝x dx 1 x ⎠ d 2 3 ⎞ ⎛2 ( J + J 32 ) = 2 ⎜ J 22 − J 32 ⎟ ⎝x dx 2 x ⎠ Adding these we get d 2 [ J + 2( J12 + J 22 + dx 0

Ch06.indd 51

)] = 0

12/13/2011 11:12:07 AM

6-52

Differential Equations

Integrating we obtain J 02 + 2( J12 + J 22 +

)=c

where c = 1 since [ J 02 ]x = 0 = J 02 (0) = 1 and J n (0) = 0 for n ≥ 1 ∴ J 02 + 2( J12 + J 20 + ) = 1 Since each term is non-negative 2

2

J 02 ≤ 1 or J 0 ≤ 1 and 2 J n ≤ 1 1 ⇒ J n ≤ for n ≥ 1 2 EXERCISE 6.2 Prove the following: 8 − x2 4 1. J 3 ( x ) = J1 ( x ) − J 0 ( x ). 2 x x ⎛ 48 8 ⎞ ⎛ 24 ⎞ 2. J 4 ( x ) = ⎜ 3 − ⎟ J1 ( x ) + ⎜1 − 2 ⎟ J 0 ( x ). ⎝x ⎠ ⎝ x x ⎠ 1 3. J 0′′ ( x ) = ( J 2 ( x ) − J 0 ( x )). 2 1 4. J 2 ( x ) = J 0′′ ( x ) − J 0′ ( x ). x J 2 ( x ) 1 J 0′′ ( x ) . = − 5. J1 ( x ) x J 0′ ( x ) d ( xJ1 ( x )) = xJ 0 ( x ). dx d n ( x J n ( ax )) = ax n J n −1 ( ax ). 7. dx 2 8. J n −1 = [nJ n − ( n + 2) J n+ 2 + ]. x x Deduce that J n = ( n + 1) J n+1 − ( n + 3) J n+ 3 − ( n + 5) J n+ 5 − 2 2sin pπ . 9. J p J −′ p − J − p J p′ = − πx ⎛ J− p ⎞ ′ 2sin pπ . Deduce that ⎜ ⎟ =− π xJ p2 ⎝ Jp ⎠ 6.

Ch06.indd 52

.

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Integration in Series

6-53

1 2 J . 2 0 a (b) ∫ xJ 0 ( λ x )dx = J1 ( aλ ).

∫J

10. (a)

0

J1dx =

λ

π

11. π J n ( x ) = ∫ cos( x sin θ − xθ )dθ . 0

12.

6.3

∫J

3

( x )dx = − J 2 ( x ) −

2 J ( x ). x 1

[JNTU 2003 (Set 2)]

CHEBYSHEV POLYNOMIALS

One of the important differential equations that gives rise to special functions is the Chebyshev differential equation (1 − x 2 ) y ′′ − xy ′ + x 2 y = 0

(6.115)

where n is a positive integer. The singularities of Eq. (6.115) are x = ±1. If we seek a power series solution of Eq. (6.115) about x = 0 of the form ∞

y( x ) = ∑ cm x m

(6.116)

m= 0

then this series solution is convergent in x < 1 since the distance between x = 0 and the nearest singularity is 1. Differentiating (6.116) w.r.t. x twice we obtain ∞

y ′( x ) = ∑ mcm x m −1

(6.117)

y ′′( x ) = ∑ m ( m − 1) cm x m − 2

(6.118)

m =1 ∞

and

m= 2

Substituting (6.116) – (6.118) in Eq. (6.115) we obtain ∞

∞

∞

∞

m= 2

m =1

m= 0

∑ m (m − 1) cm x m−2 − ∑ m( m − 1) cm x m − ∑ mcm x m + n2 ∑ cm x m = 0

m= 2

∞

or

∞

2c2 + 6c3 x + ∑ m ( m − 1) cm x m − 2 − ∑ m ( m − 1) cm x m − c1 x m= 4

m= 2

∞

− ∑ mcm x + n (c0 + c1 x ) + n2 ∑ cm x m = 0 m

m= 2

Ch06.indd 53

2

m= 2

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6-54

Differential Equations

Substituting m − 2 = t in the first sum, we get ∞

(2c2 + n2 c0 ) + (6c3 − c1 + n2 c1 ) x + ∑ (t + 2)(t + 1) ct + 2 x t + 2 t =2

∞

− ∑ [m( m − 1) + m − n2 ] cm x m = 0 m= 2

Since t is a dummy variable, we can combine the third and fourth terms of this equation. Equating the coefficients of various powers of x to zero, we obtain 2c2 + n2 c0 = 0, 6c3 − c1 + n2 c1 = 0 and ( m + 2)( m + 1) cm+ 2 − ( m2 − n2 ) cm = 0 ∴

m = 2, 3,

n2 1 − n2 n2 − 1 c0 , c3 = c1 = − c 2! 3! 3! 1 n 2 − m2 =− c , m≥2 ( m + 2)( m + 1) m

c2 = − cm+ 2

(6.119)

We have c4 = −

( n2 − 22 ) 1 c2 = n2 ( n2 − 22 ) c0 4⋅3 4!

c5 = −

( n3 − 32 ) 1 c3 = ( n2 − 12 )( n2 − 32 )c1 , 5⋅ 4 5!

Substituting in the power series solution, we obtain 1 ⎡ 1 ⎤ y( x ) = c0 ⎢1 − n2 x 2 + ( n2 − 22 )n2 x 4 + ⎥ 2! 4! ⎣ ⎦ 1 2 2 3 1 2 ⎡ + c1 ⎢ x − ( x − 1 ) x + ( n − 32 )( n2 − 12 ) x 5 + 3! 5! ⎣ = c0 y0 ( x ) + c1 y1 ( x )

⎤ ⎥⎦ (6.120)

where 1 2 2 1 2 n x + ( x − 22 ) n2 x 4 2! 4! 1 2 2 3 1 2 y1 ( x ) = x − ( x − 1 ) x + ( n − 32 )( n2 − 12 ) x 5 3! 5! y0 ( x ) = 1 −

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Integration in Series

6-55

The series y0 ( x ), y1 ( x ) converge for x < 1. Now, y0 ( x ) contains even powers of x only and y1 ( x ) contains odd powers of x only. Hence, the two solutions y0 ( x ), y1 ( x ) are linearly independent solutions of the Chebyshev differential equation (6.115). As n takes the value zero and even positive integral values we obtain for n = 0 : y0 ( x ) = 1, n = 2 : y0 ( x ) = 1 − 2 x 2 , n = 4 : y0 ( x ) = 1 − 8 x 2 + 8 x 4 , Hence, y0 ( x ) reduces to an even degree polynomial as n takes even positive integral values, whereas y1 ( x ) remains an infinite series. As n takes odd positive integral values, we get for n = 1 : y1 ( x ) = x 1 n = 3 : y1 ( x ) = (3 x − 4 x 3 ) 3 1 n = 5 : y1 ( x ) = (5 x − 20 x 3 + 16 x 5 ), 5 Hence, y1 ( x ) reduces to an odd degree polynomial as n takes odd positive integral values, whereas y0 ( x ) remains an infinite series. These polynomial give rise to an important class of polynomials called Chebyshev polynomials. Changing the independent variable x by the substitution x = cos θ we have dy dy dθ 1 dy d 2 y d ⎛ 1 dy ⎞ dθ = ⋅ =− = ⎜⎝ − ⎟ 2 dx dθ dx sin θ dθ dx dθ sin θ dθ ⎠ dx 1 ⎞ ⎡ 1 d 2 y cos θ dy ⎤ ⎛ = ⎢− + 2 ⎜⎝ − ⎟ ⎥ 2 sin θ dθ ⎦ sin θ ⎠ ⎣ sin θ dθ 2 1 d y cos θ dy = − sin 2 θ dθ 2 sin 3 θ dθ Substituting in the differential Eq. (6.115) we obtain 1 dy ⎞ ⎛ 1 d 2 y cos θ dy ⎞ ⎛ − − cos θ ⎜ − + x2 y = 0 sin 2 θ ⎜ 2 ⎝ sin θ dθ ⎟⎠ ⎝ sin θ dθ 2 sin 3 θ dθ ⎟⎠ d2 y (6.121) ⇒ + x2 y = 0 dθ 2

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6-56

Differential Equations

The general solution of the differential Eq. (6.121) is y(θ ) = A cos nθ + B sin nθ ∴ The solution of Eq. (6.115) can be written as y( x ) = A cos( n cos −1 x ) + B sin ( n cos −1 x )

(6.122)

Thus, cos( n cos −1 x ) and sin ( n cos −1 x ) are two linearly independent solutions of Eq. (6.119). Denoting the first solution by Tn ( x ) we have Tn ( x ) = cos( n cos −1 x )

−1 ≤ x ≤ 1

(6.123)

which is called the Chebyshev polynomial of first kind. Denoting the second linearly independent solution by U n ( x ) = sin ( n cos −1 x )

−1 ≤ x ≤ 1

(6.124)

which is called the Chebyshev polynomial of second kind. Chebyshev polynomials of first kind The Chebyshev polynomials of first kind are given by Tn ( x ) = cos nθ = cos( n cos −1 x ), −1 ≤ x ≤ 1

(6.125)

we note that Tn ( x ) ≤ 1. Recurrence relation for Chebyshev polynomials T n ( x ) The Chebyshev polynomials Tn ( x ) satisfy the following recurrence relation Tn+1 ( x ) − 2 xTn ( x ) + Tn −1 ( x ) = 0

(6.126)

we have Tn+1 ( x ) = cos[( n + 1) cos −1 x ] = cos ( n + 1)θ Tn −1 ( x ) = cos[( n − 1) cos −1 x ] = cos ( n − 1)θ

where cos −1 x = θ . Hence Tn+1 ( x ) = cos( nθ + θ ) = cos nθ cos θ − sin nθ sin θ and Tn −1 ( x ) = cos( nθ − θ ) = cos nθ cos θ + sin nθ sin θ Adding we obtain Tn+1 ( x ) + Tn −1 ( x ) = 2cos nθ cos θ = 2 xTn ( x ) ⇒ Tn+1 ( x ) − 2 xJ n ( x ) + J n −1 ( x ) = 0

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Integration in Series

6-57

From Eq. (6.125) we have T0 ( x ) = cos θ = 1, T1 ( x ) = cos(cos −1 x ) = x Using the recurrence relation (Eq. (6.126)) we have T2 ( x ) = 2 x T1 ( x ) − T0 ( x ) = 2 x 2 − 1 T3 ( x ) = 2 x T2 ( x ) − T1 ( x ) = 2 x(2 x 2 − 1) − x = 4 x 3 − 3x T4 ( x ) = 2 x T3 ( x ) − T2 ( x ) = 2 x (4 x 3 − 3x ) − (2 x 2 − 1) = 8 x 4 − 8 x 2 + 1 T5 ( x ) = 2 x T4 ( x ) − T3 ( x ) = 2 x (8 x 4 − 8 x 2 + 1) − (4 x 3 − 3 x ) (6.127) = 16 x 5 − 20 x 3 + 5 x ∴ Tn ( x ) is a polynomial of degree n. If n is even Tn ( x ) is an even degree polynomial and if n is odd Tn ( x ) is an odd degree polynomial. To express xn in terms of Chebyshev polynomials From Eqs. (6.127) we obtain 1 = T0 ,

x = T1 ,

1 1 x 2 = T0 + T2 2 2

x3 =

1 (3T + T ) 4 1 3

1⎡ 1 ⎤ 1 x 4 = ⎢T4 + 8 × (T0 + T2 ) − T0 ⎥ = (3T0 + 4T2 + T4 ) 8⎣ 2 ⎦ 8 1 1 ⎡ ⎤ 1 x 5 = ⎢T5 + 20 × (3T1 + T3 ) − 5T1 ⎥ = (10T1 + 5T3 + T5 ) (6.128) 16 ⎣ 4 ⎦ 16 T0

(0, 1) T1

(–1, 0)

0

(1, 0)

T3 T2 (0, –1)

Figure 6.3 Chebyshev polynomials

Ch06.indd 57

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6-58

Differential Equations

The relationship between Tn′( x ) and Tn ( x ) is given by the recurrence relation (6.129) (1 − x 2 ) Tn′( x ) = − nxTn ( x ) + nTn −1 ( x ) Putting x = cos θ we get d dθ cos nθ dθ dx 1 ⎞ ⎛ = sin 2 θ ( − n sin nθ ) ⎜ − = n sin θ sin nθ ⎝ sin θ ⎟⎠

LHS = (1 − cos2 θ )

RHS = n cos( n − 1)θ − n cos θ cos nθ = n [cos nθ cos θ + sin nθ sin θ − cos θ cos nθ ] = n sin θ sin nθ Since LHS = RHS Eq. (6.115) is proved. Example 6.3.1 Write the following polynomials in terms of Chebyshev polynomials of first kind: (a) 2 x 2 − 5 x + 7 (b) 8 x 3 + 11x 2 − 3 x + 4 (c) 16 x 4 − 8 x 3 − 2 x 2 + 4 x − 3 1 1 Solution We have 1 = T0 , x = T1 , x 2 = (T0 + T2 ), x 3 = (3T1 + T3 ), 2 4 1 4 x = (3T0 + 4T2 + T4 ) 8 1 2 x 2 − 5 x + 7 = 2 × (T0 + T2 ) − 5T1 + 7T0 = T2 − 5T1 + 8T0 2 1 11 8 x 3 + 11x 2 − 3 x + 4 = 8 × (3T1 + T3 ) + (T0 + T2 ) − 3T1 + 4T0 4 2 11 19 = 2T3 + T2 + 3T1 + T0 2 2 1 1 16 x 4 − 8 x 3 − 2 x 2 + 4 x − 3 = 16 × (3T0 + 4T2 + T4 ) − 8 × (3T1 + T0 ) 8 4 1 −2 × (T0 + T2 ) + 4T1 − 3T0 = 2T4 + 7T2 − 2T1 2

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Integration in Series

6-59

Example 6.3.2 Find the expression for (a) T3′( x ) in terms of T3 ( x ) and T2 ( x ); and (b) T5′( x ) in terms of T5 ( x ) and T4 ( x ). Solution Consider (1 − x 2 ) Tn′( x ) = − nxTn ( x ) + nTn −1 ( x ) (a) Taking n = 3 we have (1 − x 2 ) T3 ( x ) = −3 xT3 ( x ) + 3T2 ( x ) 3 ⎡T ( x ) − xT3 ( x ) ⎤⎦ ⇒ T3′( x ) = 1 − x2 ⎣ 2 (b) Taking n = 5 we have (1 − x 2 ) T5′( x ) = −5 xT5 ( x ) + 5 T4 ( x ) 5 ⎡T ( x ) − xT5 ( x ) ⎤⎦ ⇒ T5′( x ) = 1 − x2 ⎣ 4 Zeros of Tn ( x ) Equating Tn ( x ) to zero we obtain Tn ( x ) = cos( n cos −1 x ) = 0 = cos(2k + 1)

π

2 k = 0,1, 2,… ( n − 1).

⇒ n cos −1 x = (2k + 1) x = cos(2k + 1)

π

π

⇒

2

2n k = 0,1, 2,…( n − 1). The n simple zeros of Tn ( x ) are given by (6.130).

(6.130)

Turning Points (Extreme values) of T n ( x ) Putting Tn′( x ) = 0 ⇒ Tn′( x ) =

n sin ( n cos −1 x )

1 − x2 ⇒ sin ( n cos −1 x ) = 0 = sin kπ ⇒

Ch06.indd 59

n cos −1 x = kπ ⇒

=0

kπ n k = 1, 2,

x = cos

( n − 1). (6.131)

12/13/2011 11:12:11 AM

6-60

Differential Equations

Tn ( x ) attains its relative maximum or minimum at ( x − 1) points given by Eq. (6.131). At these points we have Tn ( xk ) = cos −1 ( n cos −1 xk ) = cos kπ = ( −1) k Also, at the end points of the interval [–1, 1] we have Tn ( −1) = cos nπ = ( −1) n and Tn (1) = cos(0) = 1. ∴ Tn ( x ) attains its maximum and minimum values at the ( n + 1) points ⎛ kπ ⎞ (6.132) xk = cos ⎜ ⎟ k = 0,1, 2,… n. ⎝ n⎠ Leading Coefficient of Tn(x) We have Tn ( x ) = cos nθ = Re ( e inθ ) = Re ( cos θ + i sin θ )

n

= Re ⎡⎣cos n θ + n c1 cos n −1 θ (i sin θ ) + n c2 cos n − 2 θ (i sin θ ) + 2

⎤ ⎦

expanding by Binomial Theorem = cos n θ − n c2 cos n − 2 θ sin 2 θ + n c4 cos n − 4 θ sin 4 θ − = x n − n c2 x n − 2 (1 − x 2 ) + n c4 x n − 4 (1 − x 2 )2 − = ( n c0 + n c2 + n c4 + ) x n + terms, containing x to lower degree = 2n −1 x n +

(6.133)

This shows that Tn ( x ) is a polynomial of degree n and its leading coefficient is 2n−1. Generating Function of Chebyshev Polynomials T n ( x ) Prove that ∞ 1 − xt = T ( x) t n ∑ 1 − 2 xt + t 2 n= 0 n

(6.134)

Proof The function on the LHS of (6.134), which is a function of two variables x and t is called the generating function of Chebyshev polynomials Tn ( x ). We have

Ch06.indd 60

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Integration in Series ∞

∞

6-61

∞

RHS = ∑ Tn ( x ) t n = ∑ cos nθ ⋅ t n = ∑ Re e inθ ⋅ t n n= 0

n= 0

∞

∞

n= 0

⎛ 1 ⎞ = Re ∑ e inθ ⋅ t n = Re ∑ ( e iθ t ) = Re ⎜ ⎝ 1 − e iθ t ⎟⎠ n= 0 n= 0 1 = Re 1 − cos θ t − i sin θ t 1 = Re [∵ cos θ = x ] (1 − xt ) − it 1 − x 2 (1 − xt ) + it 1 − x 2 (1 − xt ) = Re = = LHS. 2 2 2 (1 − xt ) + t (1 − x ) 1 − 2 xt + t 2 n

Hence the result (6.134) is proved. Alternate expansion for Tn(x) Writing (1 − 2 xt + t 2 ) −1 = [1 − t (2 x − t ) ] using Binomial Theorem we get

−1

and expanding it by

(1 − 2 xt + t 2 ) −1 = [1 − t (2 x − t ) ] = 1 + t (2 x − t ) + t 2 (2 x − t )2 + t n − r (2 x − t ) + −1

Now, the term containing t n in the expansion of the product t (2 x − t ) n − r is obtained from n−r

t n − r ⎡⎣ n − r cr ( −t ) r (2 x ) n − 2 r ⎤⎦ = ( −1) r

n−r

cr (2 x ) n − 2 r t n

∴ The coefficient of t n from all the terms is bn where ∞

bn = ∑ ( −1) r

n−r

r=0

cr (2 x ) n − 2 r

Also, the term containing t n −1 in the expansion of the product t (2 x − t ) n − r is obtained from n−r

t n − r ⎡⎣ n − r cr −1 ( −t ) r −1 (2 x ) n − 2 r +1 ⎤⎦ = ( −1) n −1

n−r

cr −1 (2 x ) n − 2 r +1 t n −1

∴ The coefficient of t n −1 from all the terms is bn–1 where ∞

bn −1 = ∑ ( −1) r −1 r =1

Ch06.indd 61

n−r

cr −1 (2 x ) n − 2 r +1

12/13/2011 11:12:13 AM

6-62

Differential Equations

Now, the LHS of Eq. (6.134) yields (1 − xt )(b0 + b1t + bn −1t n −1 + bnt n + ) = (b1 − bn x ) t + + (bn − x bn −1 ) t n + Thus, we obtain the coefficient of t n as ∞

bn − xbn −1 = ∑ ( −1) r ∞

r=0

− ∑ ( −1) r −1 r =1

n−r

n−r

cr (2 x ) n − 2 r

cr −1 2n − 2 r +1 x n − 2 r + 2

In the second sum we put r − 1 = s so that we have, replacing the dummy variable s by r: ∞ 1 ∞ bn − x bn −1 = ∑ ( −1) r n − r cr (2 x ) n − 2 r − ∑ ( −1) r n − r −1cr ⋅ (2 x ) n − 2 r 2 r=0 r=0 1⎡ n n n−2 = ⎣( 2 c0 − n −1 c0 ) ( 2 x ) − (2 n −1c1 − n − 2 c1 ) ( 2 x ) + 2 n−2r +2 ( n − r cr − n − r −1cr ) ( 2 x ) + ⎤⎦ consequently Tn ( x ) =

1⎡ n−2 (2 x ) n − (2n −1 c1 − n − 2 c1 ) ( 2 x ) ⎣ 2 n−4 + ( 2 n − 2 c2 − n −3c3 ) ( 2 x ) − ⎤⎦

We can also obtain the expression for x n in terms of the Chebyshev polynomials as follows: 1 n x n = (cos θ ) n = n ( e iθ + e − iθ ) 2 1 = n ⎡⎣e inθ + n c1e i ( n −1)θ e − iθ + n c2 e i ( n − 2)θ e −2iθ + 2 + n cn −1e iθ ⋅ e − i ( n −1)θ + e − inθ ⎤⎦ 1 ⎡⎛ e inθ + e − inθ ⎞ n ⎛ e i ( n − 2)θ + e − i ( n − 2)θ ⎞ = n −1 ⎢⎜ ⎟⎠ + c1 ⎜⎝ ⎟⎠ + 2 ⎣⎝ 2 2 ⎤ ⎛ e i ( n − 2 r )θ + e − i ( n − 2 r )θ ⎞ + n cn ⎜ + + ⎥ ⎟ ⎝ ⎠ 2 ⎦ collecting the equidistant terms from both ends 1 = n −1 ⎡⎣cos nθ + n c1 cos( n − 2)θ + + n cr cos( n − 2r )θ + ⎤⎦ 2 1 = n −1 ⎡⎣Tn ( x ) + n c1 Tn − 2 ( x ) + + n cn Tn − 2 r ( x ) + ⎤⎦ 2

Ch06.indd 62

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Integration in Series

If n is even, the last term is

6-63

1n . c T ( x ). 2 n2 0

Integration of Tn(x) Consider

∫ T ( x) dx = ∫ cos nθ ⋅ d (cos θ ) = − ∫ cos nθ sin θ dθ 1 = − ∫ [sin ( n + 1)θ − sin ( n − 1)θ ] dθ 2 n

1 ⎡ cos( n + 1) θ cos( n − 1) θ ⎤ n >1 − n − 1 ⎥⎦ 2 ⎢⎣ n + 1 1⎡ 1 1 ⎤ = ⎢ ⋅ Tn+1 ( x ) − ⋅ Tn −1 ( x ) ⎥ n −1 2 ⎣n +1 ⎦ =

The above result does not hold good for n = 0 and 1. For these values of n we have

∫ T ( x) dx = ∫ 1⋅ dx = x = T ( x) cos θ 1 x ∫ T ( x) dx = ∫ x dx = 2 = 2 = 4 (2cos 0

1

2

2

1

=

2

1 4

θ ) = (1 + cos 2θ )

1 (T + T2 ). 4 0

Orthogonality of Chebyshev Polynomials Tn(x) Prove that ⎧ ⎪0, if m ≠ n 1 T ( x) T ( x) ⎪ m n ∫−1 1 − x 2 dx = ⎨π , if m = n = 0 ⎪π ⎪ , if m = n ≠ 0 ⎩2 Solution Case (i) m = n = 0. In this case, we get (since T0 = 1 ) I =∫

1

−1

1 π ⎛ π⎞ 1 dx = s in −1 x −1 = sin −1 1 − sin −1 ( −1) = − ⎜ − ⎟ = π 2 2 ⎝ 2⎠ 1− x

Case (ii) m = n ≠ 0. In this case, we have to evaluate I =∫

1

−1

Ch06.indd 63

Tn2 ( x ) 1 − x2

cos 2 ( n cos −1 x ) dx −1 1 − x2

dx = ∫

1

12/13/2011 11:12:13 AM

6-64

Differential Equations

Put cos −1 x = θ or x = cos θ ⇒ dx = − sin θ dθ and the limits x = −1 and x = 1 become θ = π and θ = 0, respectively. ∴

cos 2 nθ 1 0 ( − sin θ )dθ = − ∫ (1 + cos 2 θ )dθ π 2 π 1 − cos 2 θ π 1 π 1⎛ 1 π ⎞ = ∫ (1 + cos 2 θ )dθ = ⎜ θ + sin 2 θ ⎟ = 0 ⎠0 2 2 2⎝ 2

I =∫

0

Case (iii) m ≠ n Since Tm ( x ) = cos( m cos −1 x ) and Tn ( x ) = cos( n cos −1 x ) we have to evaluate I =∫

1

cos( m cos −1 x ) cos( n cos −1 x ) 1 − x2

−1

dx

Put cos −1 x = 0 ⇒ x = cos θ , dx = − sin θ dθ = 1 − x 2 = sin θ and the limits x = −1 and x = 1 become θ = π and θ = 0, respectively. ∴ I = −∫

0

π

cos mθ cos nθ (sin θ ) dθ sin θ

1 π = ∫ [ cos( m + n)θ + cos( m − n)θ ] dθ 2 0 π ⎡ sin ( m + n)θ sin ( m − n)θ ⎤ =⎢ + m≠n m − n ⎥⎦ 0 ⎣ m+ n =0 consequently the Chebyshev polynomials Tn ( x ) are orthogonal w.r.t. the weight function w ( x ) = 1−1 x2 on [ −1,1]. Chebyshev Series Let f ( x ) be a continuous function, having continuous derivatives on the interval [–1, 1]. Now, f ( x ) can be uniquely written as an infinite series. ∞

f ( x ) = ∑ arTr ( x ) = a0T0 ( x ) + a1T1 ( x ) + a2T2 ( x ) + a3T3 ( x ) +

(6.135)

r=0

which converges uniformly in [–1, 1]. Multiplying both sides of Tm ( x )

the equation by 1− x2 and integrating with respect to x over [–1, 1] and using the orthogonal property of Chebyshev polynomials we get

Ch06.indd 64

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Integration in Series

∫

1

−1

f ( x)

1 − x2

Hence c0 =

1

π∫

1

−1

∞

Tm ( x )

dx = ∑ an ∫

1

−1

n= 0

⎧π a0 ⎪ = ⎨π ⎪⎩ 2 cn

6-65

Tm ( x )Tn ( x )

dx 1 − x2 for m = n = 0 for m = n ≠ 0.

f ( x) 2 1 f ( x ) Tm ( x ) dx, cm = ∫ dx m = 1, 2, 3, 2 π −1 1 − x 2 1− x

(6.136)

(6.137)

If we write the series given in Eq. (6.135) as 1 f ( x ) = c0T0 ( x ) + c1T1 ( x ) + c2T2 ( x ) + 2 then using Eq. (6.137) we can write cm =

2

1

f ( x ) Tm ( x )

−1

1 − x2

π∫

dx m = 0,1, 2,

(6.138)

This series may also be written as ∞

f ( x ) = ∑ ′ cnTn ( x ) n= 0

where ∑ ′ means that the coefficient of the first term T0 ( x ) is multiplied by 1/2. Example 6.3.3 Expand f ( x ) = x 3 + x, − 1 ≤ x ≤ 1 in a Chebyshev series. Solution We have f ( x) = x3 + x write f ( x ) = c0T0 ( x ) + c1T1 ( x ) + c2T2 ( x ) + we obtain c0 =

1

1

π ∫−1

f ( x) 1 1 x3 + x dx = ∫ dx = 0 π −1 1 − x 2 1 − x2

(odd function) cm =

Ch06.indd 65

2

1

π ∫−1

( x 3 + x )Tm ( x ) 1 − x2

dx = 0

for m > 3

12/13/2011 11:12:15 AM

6-66

Differential Equations

c1 =

2

1

( x 3 + x )T1 ( x )

c2 =

2

π∫

1

( x 3 + x )(2 x 2 − 1)

c3 =

2

π∫

1

4

π 2

4

π 2

x4 + x2 dx π ∫−1 π ∫−1 1 − x 2 1 − x2 4 1 x4 + x2 4 π 2 cos 4θ + cos 2 θ = ∫ dx = ∫ sin θ dθ π 0 1 − x2 π 0 sin θ 4 ⎛3 1 π 1 π⎞ 7 = ⎜ ⋅ ⋅ + ⋅ ⎟= π ⎝ 4 2 2 2 2⎠ 4

=

−1

2

1

dx = 0 (odd function) 1 − x2 ( x 3 + x )(4 x 3 − 3 x ) 1 − x2

−1

π ∫0

(cos3 θ + cos θ )(4 cos3 θ − 3cos θ ) dθ

π∫

(4 cos6 θ + cos 4 θ − 3cos 2 θ )dθ

= =

dx =

0

4 ⎡ ⎛5 3 1 π⎞ 3 1 π ⎛1 π⎞⎤ 1 4 ⎜ ⋅ ⋅ ⋅ ⎟ + ⋅ ⋅ − 3⎜ ⋅ ⎟ ⎥ = ⎢ ⎝ 2 2⎠⎦ 4 π ⎣ ⎝ 6 4 2 2⎠ 4 2 2

Hence f ( x ) =

1 (T + 7T1 ). 4 3

EXERCISE 6.3 Prove the following (1–7): 1. Tn (1) = 1.

2. Tn ( −1) = ( −1) n.

3. Tn ( − x ) = ( −1) n Tn ( x ).

4. T2 n (0) = ( −1) n.

5. T2 n+1 (0) = 0 .

6.

7.

m

1

x Tn ( x )

−1

1 − x2

∫

8. Show that

Ch06.indd 66

dx = 0

∫

1

−1

Tn ( x ) 1 − x2

dx = 0.

( m < n).

∞ 1− t2 = T ( x ) + 2 Tn ( x ) t n. ∑ 0 1 − 2 xt + t 2 n =1

12/13/2011 11:12:15 AM

7 Fourier Integral Transforms 7.1

INTRODUCTION

Integral transforms are useful in solving initial and boundary value problems and in evaluating certain integrals. Laplace and Fourier transforms are two important transforms which are widely used in engineering and physical applications. They are used in the solution of conduction of heat, vibration of strings, oscillations of elastic beams, transmission lines, etc. Here we define Fourier transform together with Fourier sine and cosine transforms, their inverses and study their properties and consider evaluation of certain integrals.

7.2

INTEGRAL TRANSFORMS

A linear integral transform or simply an integral transform of a function f (x) is defined by b

T { f ( x )} = ∫ f ( x ) K ( s, x ) dx a

(7.1)

where K(s, x), called the Kernel of the transformation, is a known function. For different choices of the Kernel K(s, x) and the limits a and b, we obtain different transforms.

7.2.1

Laplace Transform

When K(s, x) = e−sx (s > 0), a = 0 and b = ∞, we obtain the Laplace transform of f (x). It is defined by ∞

F ( s) = L{ f ( x )} = ∫ f ( x )e − sx dx ( s > 0) 0

Ch07.indd 1

(7.2)

12/9/2011 12:10:49 PM

7-2

7.2.2

Differential Equations

Fourier Transform

When K(s, x) = eisx, a = −∞ and b = ∞, we obtain the Fourier transform of f (x). It is defined by F ( s) = F { f ( x )} = ∫

∞

−∞

7.3

f ( x )e isx dx

(7.3)

FOURIER INTEGRAL THEOREM

Let f (x) be a function satisfying the following Dirichlet conditions in every interval (−l, l) (l > 0) however large: 1. f (x) is periodic; f (x) and its integrals are single-valued and finite (bounded). 2. f (x) has at most a finite number of discontinuities in any one period. 3. f (x) has at most a finite number of maxima or minima. Then the Fourier series expansion of f (x) in (−l, l) is given by f ( x) =

a0 2

∞

+ ∑ an cos n =1

npx ∞ npx + ∑ bn sin l l n =1

(7.4)

where ( a0 , an , bn ) =

1 l npt npt ⎞ ⎛ f (t ) ⎜1, cos , sin ⎟ dt ∫ − l ⎝ l l l ⎠

(7.5)

Substituting for a0, an, bn from (7.5), Eq. (7.4) becomes 1 l 1∞ l npt npx f (t )dt + ∑ ∫ cos cos f (t ) dt ∫ l l − − 2l l n=1 l l 1 ∞ l npt npx + ∑ ∫ sin sin f (t ) dt l n=1 − l l l 1 l 1∞ l np (t − x ) = ∫ f (t )dt + ∑ ∫ cos f (t ) dt 2l − l l n =1 − l l

f ( x) =

(7.6)

If we further assume that f (x) is absolutely integrable for all x i.e.,

Ch07.indd 2

∫

∞

−∞

f ( x ) dx

12/9/2011 12:10:50 PM

Fourier Integral Transforms

7-3

is convergent, then

1 l 2l ∫− l

1 l ⎤ f (t ) dt → 0 as l → ∞ ⎥ 2l ∫− l ⎥ 1 l f (t ) dt ≤ ∫ f (t ) dt → 0 as l → ∞ ⎥ ⎥⎦ 2l − l

(7.7)

Now (7.6) becomes f ( x) =

Putting

1 ∞ l np (t − x ) cos f (t ) dt ∑ ∫ l − l n= 0 l

p 1 ds = ds or = we have, on taking limit as n → ∞ or ds → 0 l l p ∞ l 1 lim ∑ ds∫ cos[nds(t − x )] f (t ) dt p n→∞ n= 0 − l 1 ∞⎡ ∞ = ∫ ∫ cos[ s(t − x ) f (t ) dt ]⎤ ⎦⎥ p 0 ⎣⎢ −∞ ∞ as lim f ( nds ) = ∫ f ( s) ds

f ( x) =

n→∞ or ds → 0

0

(7.8)

which is called the Fourier Integral (FI) of f (x). Note 7.3.1 A rigorous proof of the above theorem is beyond the scope of this book. Note 7.3.2 Fourier integral representation for a function f (x) is useful in solving certain differential and integral equations.

7.3.1 Fourier Sine and Cosine Integrals (FSI’s and FCI’s) Writing the expansion for cosine in (7.8) we have 1 ∞ ∞ (cos st cos sx + sin st sin sx ) f (t ) dt ds p ∫0 ∫−∞ 1 ∞ ⎛ ∞ ⎞ = ∫ cos sx ⎜ ∫ cos st f (t ) dt ⎟ ds ⎝ −∞ ⎠ p 0 1 ∞ ⎛ ∞ ⎞ + ∫ sin sx ⎜ ∫ sin st f (t ) dt ⎟ ds −∞ 0 ⎝ ⎠ p

f ( x) =

Ch07.indd 3

(7.9)

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7-4

Differential Equations

When f (t) is an odd function in (−∞, ∞) then the integral in the first brackets in (7.9) is zero since the integrand cos st f (t) is an odd function in (−∞, ∞) and so we obtain f ( x) =

2 ∞ ⎛ ∞ ⎞ sin sx ⎜ ∫ sin st f (t ) dt ⎟ ds ∫ 0 0 ⎝ ⎠ p

(7.10)

which is called the Fourier Sine Integral (FSI) of f (x). When f (t) is an even function in (−∞, ∞) then the integral in the second brackets in (7.9) is zero since the integrand sin st f (t) is an odd function in (−∞, ∞) and so we obtain f ( x) =

⎛ ∞ ⎞ 2 ∞ cos sx ⎜ ∫ cos st f (t ) dt ⎟ ds ∫ 0 0 ⎝ ⎠ p

(7.11)

which is called the Fourier Cosine Integral (FCI) of f (x).

7.4

FOURIER INTEGRAL IN COMPLEX FORM

cos s(t − x) is an even function of s in (−∞, ∞) and so we can write (7.8) in the form 1 ∞ ∞ f ( x) = cos[ s(t − x )] f (t ) dt ds (7.12) 2p ∫−∞ ∫−∞ Also, sin s(t − x) is an odd function of s in (−∞, ∞). So,we can write 0=

1 2p

∞

∫ ∫

∞

−∞ −∞

sin[ s(t − x )] f (t ) dt ds

(7.13)

Now, multiplying (7.13) by ‘i ’ and adding it to (7.12) we obtain f ( x) =

1 ∞ ∞ {cos[ s(t − x )] 2p ∫−∞ ∫∞ + i sin[ s(t − x )]} f (t ) dt ds

Using Euler’s formula eiq = cos q + i sin q we can write the above integral as 1 ∞ ∞ (7.14) f ( x) = f (t )e is ( t − x ) dt ds 2p ∫−∞ ∫−∞ which is called Fourier Integral of f (x) in complex form (FICF )

7.4.1 Fourier Integral Representation of a Function By (7.9) a function f (x) may be represented by a FI as f ( x) =

Ch07.indd 4

1 ∞ [ A( s) cos sx + B( s)sin sx ] ds p ∫0

(7.15)

12/9/2011 12:10:52 PM

Fourier Integral Transforms ∞

where

A( s) = ∫

and

B( s) = ∫

−∞ ∞

−∞

7-5

f (t ) cos st dt

(7.16)

f (t )sin st dt

(7.17)

If f (x) is an odd function in (−∞, ∞) then 1 ∞ B( s)sin sx ds p ∫0

f ( x) =

(7.18)

where ∞

B( s) = 2∫ f (t )sin st dt 0

(7.19)

and if f (x) is an even function in (−∞, ∞) then f ( x) =

where

1 ∞ A( s) cos sx ds p ∫0 ∞

A( s) = 2∫ f (t ) cos st dt 0

7.5

(7.20)

(7.21)

FOURIER TRANSFORM OF F X

Writing the exponential function eix (t − x) as a product of two exponential functions eix (t − x) = eist · e−isx the FI (7.14) may be put in the following form f ( x) =

1 2p

⎛ ∞ ⎞ e − isx ⎜ ∫ f (t )e ist dt ⎟ ds −∞ ⎝ −∞ ⎠

∫

∞

(7.22)

The expression inside the brackets is a function of s. Denoting it by F (s) and replacing t by x, we have the Fourier transform (FT) of f (x). Then the inverse Fourier transform (IFT) of F(s) is given by (7.22). Fourier Transform The Fourier transformation of f (x), denoted by F{ f (x)}, is defined by F { f ( x )} = F ( s) = ∫

∞

f ( x )e isx dx

−∞

(7.23)

Inverse Fourier Transform The inverse FT of F(s), denoted by F −1 {F (s)} is defined by F −1{F ( s)} = f ( x ) =

Ch07.indd 5

1 2p

∫

∞

−∞

f ( s)e − isx ds

(7.24)

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7-6

Differential Equations

Thus, the function F (s) defined by (7.23) is called the Fourier Transform of f (x) and is denoted by F{ f (x)} = F (s)

(7.25)

and the function f (x) given by (7.24) is called the Inverse Fourier transform of F (s) and is denoted by F −1{F (s)} = f (x)

(7.26)

The process of obtaining the FT F{f (x)} = F (s) from a given function f (x) is called the FT method or simply Fourier Transform. Existence of Fourier Transform The following conditions are sufficient for the existence of the FT of a function f (x): 1. f (x) is piecewise continuous on every finite interval. 2. f (x) is absolutely integrable for all x.

7.5.1

Fourier Sine Transform (FST) and Fourier Cosine Transform (FCT)

The FSI in (7.10) can be written by replacing t by x as ∞

f ( x ) = ∫ sin sx 0

(∫

∞

0

)

f ( x )sin sx dx ds

(7.27)

The expression in the brackets in (7.27) is a function of s denoted by Fs(s). Writing ∞

Fs { f ( x )} = Fs ( s) = ∫ f ( x )sin sx dx 0

(7.28)

Equation (7.20) becomes Fs−1{Fs ( s)} = f ( x ) =

2 ∞ F ( s)sin sx ds p ∫0 s

(7.29)

The function defined by (7.28) is called the Fourier sine transform (FST) of f (x) and that defined by (7.29) is called the inverse Fourier sine transform of Fs(s). The Fourier cosine integral (7.11) can be written by replacing t by x as ∞ ⎛ ∞ ⎞ f ( x ) = ∫ cos sx ⎜ ∫ f ( x ) cos sx dx ⎟ ds 0 ⎝ 0 ⎠

Ch07.indd 6

(7.30)

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Fourier Integral Transforms

7-7

The expression in the brackets (7.30) is a function of s denoted by Fc(s). Writing ∞

Fc { f ( x )} = Fc ( s) = ∫ f ( x ) cos sx dx 0

Fc −1{Fc ( s)} = f ( x ) =

2 ∞ F ( s) cos sx dx p ∫0 c

(7.31) (7.32)

The functions Fc(s) and Fc−1 {Fc(s)} = f (x) are, respectively, called the Fourier cosine transform and inverse Fourier cosine transform, respectively.

7.6

FINITE FOURIER SINE TRANSFORM AND FINITE FOURIER COSINE TRANSFORM FFCT

The finite Fourier sine transform of f (x) in 0 < x < l is defined by l

Fs ( n) = ∫ f ( x )sin 0

npx dx l

(7.33)

where n is an integer. The function f (x) is then called the inverse finite Fourier sine transform of Fs(n) and is given by f ( x) =

2 ∞ npx Fs ( n)sin ∑ l n =1 l

(7.34)

Similarly the finite Fourier cosine transform of f (x) in 0 < x < l is defined by l

Fc ( n) = ∫ f ( x ) cos 0

npx dx l

(7.35)

where n is an integer. The function f (x) is then called the inverse finite Fourier cosine transform of Fc(n) and is given by 1 2 ∞ npx f ( x ) = Fc (0) + ∑ Fc ( n) cos l l n =1 l

(7.36)

Note 7.6.1 The finite Fourier sine transform is useful for solving problems with boundary condition of heat distribution (say) on two parallel boundaries while the finite Fourier cosine transform is useful for solving problems in which the velocity distribution (say) normal to two parallel boundaries are prescribed.

Ch07.indd 7

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7-8

Differential Equations

7.6.1 FT, FST and FCT Alternative definitions The Fourier transform, FST and FCT are alternatively defined as follows: Definition 7.6.2 1 2p

∫

∞

FT

F ( s) =

IFT

f ( x ) = ∫ F ( s)e − isx ds

FST IFST

FCT IFCT

−∞

f ( x )e isx dx,

∞

−∞

2 ∞ f ( x )sin sx dx, p ∫0 ∞ f ( x ) = ∫ Fs ( s)sin sx ds

(7.37)

Fs ( s) =

0

2 ∞ f ( x ) cos sx dx, p ∫0 ∞ f ( x ) = ∫ Fc ( s) cos sx ds

(7.38)

Fc ( s) =

(7.39)

0

Definition 7.6.3 FT IFT FST IFST

FCT IFCT

1 2p 1 f ( x) = 2p

F ( s) =

∫

∞

∫

∞

−∞

−∞

f ( x )e isx dx , F ( s)e − isx ds

2 ∞ f ( x )sin sx dx p ∫0 2 ∞ f ( x) = F ( s)sin sx ds p ∫0 s

(7.40)

Fs ( s) =

2 ∞ f ( x ) cos sx dx p ∫0 2 ∞ f ( x) = F ( s) cos sx ds p ∫0 c

(7.41)

Fc ( s) =

(7.42)

Note 7.6.4 Each of these notations has its advantages and disadvantages. So, one can adopt any one of these definitions. If both transformation and inverse transformation are involved completing one cycle there will not be any difference in the results. If only one of these is

Ch07.indd 8

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Fourier Integral Transforms

7-9

involved the results will be different. The student, while answering a question, should understand which notation is used and then answer, clearly stating the notation he or she is using.

7.7

CONVOLUTION THEOREM FOR FOURIER TRANSFORMS 7.7.1 Convolution Definition 7.6.5 The convolution of two functions f (x) and g (x) over the interval (−∞, ∞) denoted by f * g is defined by h ( x) = f ∗ g = ∫

∞

−∞

7.7.2

f (u ) g ( x − u ) dx

(7.43)

Convolution Theorem

Theorem 7.6.6 If F (s) = F{ f (x)} and G(s) = F{g (x)} are the Fourier transforms of f (x) and g (x) then the Fourier transform of the convolution of f (x) and g (x) is the product of their transforms F{ f (x) * g (x)} = F{ f (x)} · F{g (x)}

(7.44)

Proof By the definition of Fourier transforms, we have F{ f ( x ) * g ( x )} = F

{∫

∞

−∞

f (u ) g ( x − u ) du

∞

∞

−∞

−∞

}

∫ ⎡⎣⎢∫ f (u) g ( x − u)du⎤⎦⎥ e dx = ∫ f (u ) ⎡ ∫ g ( x − u )e dx ⎤ du, ⎢⎣ ⎥⎦

=

isx

∞

∞

−∞

−∞

isx

Changing the order of integration ∞ ∞ = ∫ f (u ) ⎡ ∫ e is ( x − u ) g ( x − u )d ( x − u ) ⎤ e isu du −∞ ⎣⎢ −∞ ⎦⎥ ∵ dx = d ( x − u ) e isx = e is ( x − u ) ⋅ e isu ∞ ∞ = ∫ f (u )e isu ⎡ ∫ g (t )e ist dt ⎤ du, −∞ ⎣⎢ −∞ ⎦⎥ where t = x − u, dt = d ( x − u ) ∞

= ∫ f (u ) ⋅ e isu du ⋅ G ( s) = F ( s) ⋅ G ( s) −∞ = F { f ( x )} ⋅ F {g ( x )}

Ch07.indd 9

12/9/2011 12:10:55 PM

7-10

Differential Equations

7.7.3

Relation between Laplace and Fourier Transforms ⎡ g (t ) e − xt, f (t ) = ⎢ ⎣ 0, F { f (t )} = L{g (t )}

If then

t>0 t 0, x > 0 .

Solution The Fourier cosine integral is given by f ( x) =

Ch07.indd 12

∞ 2 ∞ cos l x ∫ f (t ) cos lt dt d l ∫ 0 0 p

(7.51)

12/9/2011 12:10:56 PM

Fourier Integral Transforms

7-13

Let f (t) = e⫺at so that f (x) = e⫺ax. Substituting in the above integral, we have

(

)

∞ ∞ 2 ∞ cos lx ∫ cos lx ∫ e − at cos lt dt d l ∫ 0 0 p 0 ∞ at − 2 ∞ ⎡ e ⎤ = ∫ cos lx ⎢ 2 − a l t + b l t ( cos sin ) ⎥⎦ d l p 0 ⎣ a + l2 0 e − a0 2 ∞ ⎡ ⎤ = ∫ cos lx ⎢0 − 2 ( − a ⋅ 1 + b ⋅ 0) ⎥ d l p 0 a + l2 ⎣ ⎦ 2a ∞ cos lx = dl p ∫0 a 2 + l 2 ∞ cos lx pe − ax ⇒ ∫ 2 d l = 0 a + l2 2a

e − ax =

(7.52)

∞ cos lx pe − x =∫ 2 d l. 0 l +1 2

Example 7.8.8 Prove that

Solution Putting a = 1 in Example 7.1, we obtain

∫

∞

0

cos lx pe − x dl = 2 2 a +l 2

⎡ ⎢ Example 7.8.9 Express f ( x ) = ⎢ ⎣⎢ integral and hence evaluate

∫

∞

0

0, − ∞ < x < −1 1, − 1 < x < 1 as a Fourier 0, 1 < x < ∞

sin s cos sx ds. s

Solution The Fourier integral of f (x) is given by f ( x) =

1 ∞ ∞ f (t ) cos s(t − x ) dt ds p ∫0 ∫−∞

(7.53)

The funtion f (t) is defined by ⎡0, f (t ) = ⎢⎢1, ⎢⎣0,

in ( −∞,1) in ( −1,1) in (1, ∞)

(7.54)

∴ We obtain

Ch07.indd 13

12/9/2011 12:10:56 PM

7-14

Differential Equations 1

1 ∞ 1 1 ∞ ⎡ sin s(t − x ) ⎤ [cos s(t − x ) dt ] ds = ∫ ⎢ ∫ ∫ ⎥⎦ ds 0 1 − p p 0 ⎣ s −1 1 ∞1 = ∫ [sin s(1 − x ) − sin s( −1 − x )] ds p 0 s 2 ∞ sin s cos sx = ∫ ds (7.55) p 0 s

f ( x) =

which is the Fourier integral of f (x). From (7.55) we obtain

∫

∞

0

⎡p , sin s cos sx p ds = f ( x ) = ⎢ 2 ⎢ 2 s ⎣0,

x ≤1

by

(7.56)

x ≥1

At x = ±1, f (x) is discontinuous and the value of the integral is 1⎛p ⎞ p ⎜⎝ + 0⎟⎠ = 2 2 4

1 [ f (1 − 0) + (1 + 0)] = 2

∫

Example 7.8.10 Prove that

∞

0

(7.57)

sin x p = . x 2

Solution From Example 7.3 ∞ sin s cos sx p ds = , ∫0 s 2

for x < 1

Putting x = 0 we have

∫

∞

0

sin x p dx = x 2

where we have replaced the dummy variable s by x. Example 7.8.11 Using the Fourier integral representation, show that ⎡0, ⎢p ∞ cos sx + s sin sx ⎢ , = ds ∫0 1 + s2 ⎢2 ⎢pe − s , ⎣

Ch07.indd 14

x 0.

12/9/2011 12:10:57 PM

Fourier Integral Transforms

Solution Let f (x) be defined by ⎡0, ⎢1 f ( x) = ⎢ , ⎢2 ⎢e − x , ⎣ We now find the Fourier integral exponential form, which is given by

7-15

( x < 0) ( x = 0)

(7.58)

( x > 0)

representation of f (x) in the

1 ⎡ ∞ f (t )e ist dt ⎤ e − isx ds ⎥⎦ 2π ⎢⎣ ∫−∞ 0 ∞ ∞ Let I = ∫ f (t )e ist dt = ∫ 0 ⋅ e ist dt + ∫ e − t .e ist dt f ( x) =

−∞

∞

−∞

0

(7.59)

0 − 1 1 + is ⎡ e − t (1−is ) ⎤ =⎢ = = ⎥ ⎣ is − 1 ⎦ 0 is − 1 1 + s 2 1 ∞ 1 + is isx 1 ∞ (1 + is)(cos sx − i sin sx ) e ds = ds ∴ f ( x) = ∫ 2 −∞ 2π 1+ s 2π ∫−∞ 1 + s2 1 ∞ cos sx + s sin sx = ds + 0, 2π ∫−∞ 1 + s2 The second integral = 0 since the integrand is odd in( − ∞, ∞) 1 ∞ cos sx + s sin sx ds, since the integrand = ∫ π 0 1 + s2 is even in ( − ∞, ∞) = e − x , for x > 0 by definition of f ( x )

{

The function f (x) is discontinuous at x = 0 and its value is 1 ∞ 1 1 f (0 + 0) + f (0 − 0) ] = (1 + 0) = [ ∫ −∞ 2 2 2 1 ∞ cos sx + s sin sx 1 ∴ ds = p ∫0 1 + s2 2 ∞ cos sx + s sin sx p ⇒ ∫ ds = , for x = 0 2 0 1+ s 2 = pe − x , for x > 0. (7.60)

f (0) =

Example 7.8.12 Express f ( x ) = ⎡⎢ Evaluate

Ch07.indd 15

1, ⎣0,

for x ≤ a as a Fourier integral. for x > a

12/9/2011 12:10:57 PM

7-16

Differential Equations ∞

(a ) ∫0

sin x dx x

∞

( b) ∫−∞

sin as cos sx ds s

[JNTU 2003(2), 2004(3)] Solution The Fourier integral of f (x) is 1 ∞ ∞ f (t ) cos s(t − x ) dt ds p ∫0 ∫−∞ 1 ∞ a 1 ∞ sin s(t − x ) = ∫ ⎡ ∫ cos s(t − x )dt ⎤ ds = ∫ ⎦⎥ p 0 ⎣⎢ − a p 0 s

f ( x) =

1 ∞ sin s( a − x ) + sin s( a + x ) ds s p ∫0 2 ∞ sin sa cos sx = ∫ ds s p 0

a

ds −a

=

⇒

∫

∞

0

⎡p .1, sin sa cos sx p ds = f ( x ) = ⎢ 2 ⎢ s 2 ⎣0,

x ≤a x >a

For |x| = a ⇔ x = ± a, which are points of discontinuity the value of the integral is p f ( a + 0) + f ( a − 0) p ⎛ 1 + 0 ⎞ p = ⋅⎜ ⎟= 2 2 2 ⎝ 2 ⎠ 4

∫

Choosing x = 0 and a = 1 we obtain

∞

0

sin x p dx = where we have x 2

replaced the dummy variable s by x.

{

1, for 0 ≤ x ≤ p

Example 7.8.13 Express the function f ( x ) = 0, for x > p Fourier sine integral. Hence, evaluate

∫

∞

0

as a

1 − cos p sin sx ds s

Solution The Fourier sine integral of f (x) is given by f ( x) =

Ch07.indd 16

2 ∞ sin sx p ∫0

(∫

∞

0

)

f (t )sin st dt ds

12/9/2011 12:10:57 PM

Fourier Integral Transforms

(

7-17

)

∞ 2 ∞ sin sx ∫ 1 ⋅ sin st dt ds ∫ 0 0 p p 2 ∞ ⎛ − cos st ⎞ = ∫ sin sx ⎜ ds ⎝ s ⎟⎠ 0 p 0 2 ∞ ⎛ − cos p + cos 0 ⎞ = ∫ sin sx ⎜ ⎟⎠ ds ⎝ p 0 s 2 ∞ sin sx (1 − cos sp ) = ∫ ds p 0 s

=

⇒

∫

∞

sin sx (1 − cos sp ) s

0

ds =

⎧p p ⎪ , 0≤ x